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    <figure class="highlight xml"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br></pre></td><td class="code"><pre><span class="line">title: LeetCode</span><br><span class="line"></span><br><span class="line">date: 2023-07-03 </span><br></pre></td></tr></table></figure>

<h1 id="刷题进度（66）"><a href="#刷题进度（66）" class="headerlink" title="刷题进度（66）"></a>刷题进度（66）</h1><span id="more"></span>

<h1 id="十大排序算法"><a href="#十大排序算法" class="headerlink" title="十大排序算法"></a>十大排序算法</h1><blockquote>
<p>稳定排序：如果 a 原本在 b 的前面，且 a&#x3D;&#x3D;b，排序之后 a 仍然在 b 的前面，则为稳定排序；</p>
<p>非稳定排序：如果 a 原本在 b 的前面，且 a&#x3D;&#x3D;b，排序之后 a 可能不在 b 的前面，则为稳定排序；</p>
<p>原地排序：指在排序过程中不申请多余的存储空间，只用原来的数组进行比较和交换的排序；</p>
<p>非原地排序：需要额外新建数组来辅助排序；</p>
<p>时间复杂度：一个算法执行所消耗的时间；</p>
<p>空间复杂度：运行一个算法所需的内存大小；</p>
</blockquote>
<h4 id="1-选择排序（交换）"><a href="#1-选择排序（交换）" class="headerlink" title="1. 选择排序（交换）"></a>1. 选择排序（交换）</h4><ul>
<li>思路：</li>
</ul>
<p>每次遍历数组，找出最小数，放到第一个位置；然后再次找剩余数组的最小的元素，放到第二个位置；</p>
<ul>
<li>程序：</li>
</ul>
<ol>
<li><p><strong>创建两个循环：</strong>一个循环用来控制所有数；一个循环用来寻找最小的数；</p>
</li>
<li><p>每次找剩余非排序数组中的最小的元素，放到有序数组中</p>
</li>
</ol>
<ul>
<li>结果：</li>
</ul>
<p>时间复杂度 O(n^2) 	空间复杂度  O(1) 	原地排序 	非稳定排序</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 遍历数组， 用来保存排序后的数组</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i&lt; nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 指向最小的数</span></span><br><span class="line">            <span class="type">int</span> min = i;</span><br><span class="line">            <span class="comment">// 从下一个数开始遍历</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = i + <span class="number">1</span>; j&lt; nums.<span class="built_in">size</span>(); j++)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 如果当前的值小于min，则赋值给min</span></span><br><span class="line">                <span class="keyword">if</span>(nums[j] &lt; nums[min])</span><br><span class="line">                &#123;</span><br><span class="line">                    min = j;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 互换</span></span><br><span class="line">            <span class="type">int</span> t = nums[min];</span><br><span class="line">            nums[min] = nums[i];</span><br><span class="line">            nums[i] = t;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 找出当前最小的数</span></span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h4 id="2-插入排序（比较）"><a href="#2-插入排序（比较）" class="headerlink" title="2. 插入排序（比较）"></a>2. 插入排序（比较）</h4><ul>
<li>思路：</li>
</ul>
<p>​	分成<strong>有序</strong>部分（数组前段）和<strong>无序</strong>部分（数组后段），从第二个数开始遍历无序部分，依次跟有序部分比较大小，比有序部分小，则将有序部分往后挪。</p>
<ul>
<li>程序：</li>
</ul>
<ol>
<li><strong>创建两个循环：</strong>一个循环用来控制所有数；一个循环用来寻找插入的位置：</li>
<li>从第二个元素开始（默认只有一个元素的时候是有序的），与左边数组最右边数开始比较，小的话该元素往左移，大的话不动；</li>
<li>（把遍历到的数拿出来，类似交换元素的操作，找到位置后再将数放进去，注意：放进去的位置一定是在小的数右边！！！）。</li>
</ol>
<ul>
<li>结果：</li>
</ul>
<p>时间复杂度 O(n^2) 	空间复杂度O(1) 	原地排序 	稳定排序</p>
<p>❗❗❗❗❗❗❗❗</p>
<p>​	条件判断的时候，重要的条件放在前面！！！！</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 1. 创建一个循环用来控制所有数进行插入排序</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">1</span>; i &lt; nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">        	<span class="comment">// 拿出这个位置的数值</span></span><br><span class="line">            <span class="type">int</span> temp = nums[i];</span><br><span class="line">            <span class="comment">// 记住这个位置的下标</span></span><br><span class="line">            <span class="type">int</span> p = i;</span><br><span class="line">            <span class="comment">// 2. 创建一个循环用来判断有序数组的数是否大于取出的数</span></span><br><span class="line">            <span class="keyword">while</span>( p &gt; <span class="number">0</span> &amp;&amp; nums[p<span class="number">-1</span>] &gt; temp )</span><br><span class="line">            &#123;</span><br><span class="line">              	<span class="comment">// 大于的话将数组右移动</span></span><br><span class="line">                nums[p] = nums[p<span class="number">-1</span>];</span><br><span class="line">                <span class="comment">// 继续判断下一位</span></span><br><span class="line">                p--;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 找到位置后放下</span></span><br><span class="line">            nums[p] = temp;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="3-冒泡排序（交换）"><a href="#3-冒泡排序（交换）" class="headerlink" title="3. 冒泡排序（交换）"></a>3. 冒泡排序（交换）</h4><ul>
<li>思路：</li>
</ul>
<p>相邻的左右两个数对比，若左边大就互换位置，直到换到最后一个位置；</p>
<ul>
<li>程序：</li>
</ul>
<ol>
<li><p><strong>创建两个循环：</strong>一个循环用来控制所有数；一个循环用来控制一轮的冒泡比较；</p>
</li>
<li><p>冒泡：从第一个元素开始，往右比，比右边元素大的就互换，直到换到最右边；</p>
</li>
<li><p>每次选数组的第一个元素进行冒泡</p>
</li>
</ol>
<ul>
<li>结果：</li>
</ul>
<p>时间复杂度 O(n^2) 	空间复杂度 O(1) 	原地排序 	稳定排序</p>
<ul>
<li><h5 id="3-1-非优化版本"><a href="#3-1-非优化版本" class="headerlink" title="3.1 非优化版本"></a>3.1 非优化版本</h5></li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="type">int</span> temp;</span><br><span class="line">        <span class="comment">// 首先判断数组是否为空，或小于2个数</span></span><br><span class="line">        <span class="keyword">if</span>(nums.<span class="built_in">size</span>() &lt;= <span class="number">1</span>) <span class="keyword">return</span> nums;</span><br><span class="line">        <span class="comment">// 1. 创建一个循环用来控制数组的所有数进行冒泡</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;	</span><br><span class="line">            <span class="comment">// 2. 创建一个循环用来控制某一个数的冒泡</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">1</span> ; j &lt; nums.<span class="built_in">size</span>() - i; j++)</span><br><span class="line">            &#123;	</span><br><span class="line">                <span class="comment">// 3. 冒泡过程（若左边的数大于右边，则交换位置）</span></span><br><span class="line">                <span class="keyword">if</span>(nums[j<span class="number">-1</span>] &gt; nums[j])</span><br><span class="line">                &#123;</span><br><span class="line">                    <span class="comment">// 交换位置</span></span><br><span class="line">                    temp = nums[j<span class="number">-1</span>];</span><br><span class="line">                    nums[j<span class="number">-1</span>] = nums[j];</span><br><span class="line">                    nums[j] = temp;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li><h5 id="3-2-优化版本"><a href="#3-2-优化版本" class="headerlink" title="3.2 优化版本"></a>3.2 优化版本</h5></li>
</ul>
<p>（添加标志位，如果是有序数组，则可以不需要继续冒泡）</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="type">bool</span> m;</span><br><span class="line">        <span class="type">int</span> temp;</span><br><span class="line">        <span class="comment">// 首先判断数组是否为空，或小于2个数</span></span><br><span class="line">        <span class="keyword">if</span>(nums.<span class="built_in">size</span>() &lt;= <span class="number">1</span>) <span class="keyword">return</span> nums;</span><br><span class="line">        <span class="comment">// 1. 创建一个循环用来控制数组的所有数进行冒泡</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 2. 创建一个判断有序数组的标志位</span></span><br><span class="line">            m = <span class="literal">true</span>;</span><br><span class="line">            <span class="comment">// 3. 创建一个循环用来控制某一个数的冒泡</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">1</span> ; j &lt; nums.<span class="built_in">size</span>() -i ; j++)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 4. 冒泡过程（若左边的数大于右边，则交换位置）</span></span><br><span class="line">                <span class="keyword">if</span>(nums[j<span class="number">-1</span>] &gt; nums[j])</span><br><span class="line">                &#123;</span><br><span class="line">                    <span class="comment">// 4.1 有冒泡说明并不是有序数组</span></span><br><span class="line">                    m = <span class="literal">false</span>;</span><br><span class="line">                    <span class="comment">// 4.2 交换位置</span></span><br><span class="line">                    temp = nums[j<span class="number">-1</span>];</span><br><span class="line">                    nums[j<span class="number">-1</span>] = nums[j];</span><br><span class="line">                    nums[j] = temp;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 5. 一个数冒泡结束后，判断有序数组标志位 （若有序数组标志位依然是true，说明遍历数组一趟下来都是有序的，可以直接停止循环）</span></span><br><span class="line">            <span class="keyword">if</span>(m == <span class="literal">true</span>) <span class="keyword">return</span> nums;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="4-希尔排序"><a href="#4-希尔排序" class="headerlink" title="4. 希尔排序"></a>4. 希尔排序</h4><ul>
<li>思路：</li>
</ul>
<p>插入排序是以1的间隔，希尔排序则是以 step 为间隔。多加一层控制步长 step的循环，从分组数组的末尾位置开始希尔排序</p>
<ul>
<li><p>程序：</p>
</li>
<li><p>结果：</p>
</li>
</ul>
<p>​	时间复杂度 O(nlogn) 	空间复杂度 O(1)</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br></pre></td><td class="code"><pre><span class="line"><span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        <span class="type">int</span> n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">if</span>(n &lt; <span class="number">1</span>) <span class="keyword">return</span> nums;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 1. 创建一个循环用来控制步长的变化（开始步长：n/2 --- 结束步长条件：step &gt;0 --- 每次变化为：n/2）</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="type">int</span> step = n/<span class="number">2</span>; step&gt;<span class="number">0</span>; step = step/<span class="number">2</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 2. 创建一个循环，用来控制所分组的末尾数字下标（开始：step 结束：数组末尾）</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> end = step; end &lt; n; end++)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 将这个数拿出来</span></span><br><span class="line">                <span class="type">int</span> temp = nums[end];</span><br><span class="line">                <span class="comment">// 记住这个位置</span></span><br><span class="line">                <span class="type">int</span> p = end;</span><br><span class="line">                <span class="keyword">while</span>(p -step &gt; <span class="number">-1</span> &amp;&amp; nums[p-step] &gt; temp)</span><br><span class="line">                &#123;</span><br><span class="line">                    <span class="comment">// 往右挪一位</span></span><br><span class="line">                    nums[p] = nums[p-step];</span><br><span class="line">                    <span class="comment">// 继续判断下一位</span></span><br><span class="line">                    p = p - step;</span><br><span class="line">                &#125;</span><br><span class="line">                <span class="comment">// 如果找到放置的位置</span></span><br><span class="line">                nums[p] = temp;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br></pre></td></tr></table></figure>

<h4 id="5-归并排序"><a href="#5-归并排序" class="headerlink" title="5. 归并排序"></a>5. 归并排序</h4><p>递归算法示意图</p>
<p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/9ZGMQ8CQ$P%60RWLNUOXI%25$EU.png"></p>
<ul>
<li><strong>思路：</strong></li>
</ul>
<p>分封而治，将数组无限二分，直至只剩下一个数；采用递归的方法分，很快；</p>
<p>递归算法：自己调用自己，一定要设置条件，什么情况下调用自己，否则函数将无限执行下去；递归算法如果有两条自身函数的代码，一定是先执行第一个，一直递归执行到最底层的递归，然后执行到底层的那个函数的第二个函数，然后再返回执行倒数第二个函数的第二条代码，直到返回执行第一个函数的第二条代码，结束；</p>
<ul>
<li><strong>程序：</strong></li>
</ul>
<ol>
<li><p>需要写一个拆分数组的函数：计算中点，将数组分为左右两组；继续计算中值，继续分；…..（递归）</p>
</li>
<li><p>需要写一个合并数组的函数：判断左右两个数组的最左边的元素，小的放进数组，下标右移动，直到所有元素遍历一遍；</p>
</li>
<li><p>需要写一个归并排序的函数：计算中值，分左右两边数组，递归，一直分到最后剩一个元素，再开始合并数组；自顶向下拆，再自底向上合；</p>
</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 法一：容易导致溢出</span></span><br><span class="line"><span class="type">int</span> mid = (start + end)/<span class="number">2</span>;</span><br><span class="line"><span class="comment">// 法二：推荐</span></span><br><span class="line"><span class="type">int</span> mid = start + (end - start)/<span class="number">2</span>;</span><br></pre></td></tr></table></figure>

<ul>
<li><strong>结果：</strong></li>
</ul>
<p>时间复杂度 O(nlogn) 	空间复杂度O(n) 	</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"><span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        <span class="type">int</span> n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">if</span>(n &lt; <span class="number">1</span>) <span class="keyword">return</span> nums;</span><br><span class="line"></span><br><span class="line">        <span class="built_in">mergesort</span>(nums, <span class="number">0</span>, nums.<span class="built_in">size</span>()<span class="number">-1</span>);</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line"><span class="comment">// 拆</span></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">mergesort</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; arr, <span class="type">int</span> start, <span class="type">int</span> end)</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">    <span class="keyword">if</span>(start &lt; end)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="comment">// 找到中间值</span></span><br><span class="line">        <span class="comment">// [分]：将数组一分为2</span></span><br><span class="line">        <span class="type">int</span> mid = start + (end - start)/<span class="number">2</span>;</span><br><span class="line">        <span class="comment">// 将左边的数组排序</span></span><br><span class="line">        <span class="built_in">mergesort</span>(arr, start, mid);</span><br><span class="line">        <span class="comment">// 将右边的数组排序</span></span><br><span class="line">        <span class="built_in">mergesort</span>(arr, mid+<span class="number">1</span>, end);</span><br><span class="line">        <span class="comment">// 合并数组</span></span><br><span class="line">        <span class="built_in">merge</span>(arr, start, mid, mid+<span class="number">1</span>, end);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line"><span class="comment">// 合</span></span><br><span class="line"><span class="function"><span class="type">void</span> <span class="title">merge</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; arr, <span class="type">int</span> start1, <span class="type">int</span> end1, <span class="type">int</span> start2, <span class="type">int</span> end2)</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">    <span class="comment">// 从数组中拿出左右两个数组的元素</span></span><br><span class="line">    <span class="type">int</span> n_left  = end1 - start1 + <span class="number">1</span>;</span><br><span class="line">    <span class="type">int</span> n_right = end2 - start2 + <span class="number">1</span>;</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">v_left</span><span class="params">(n_left)</span></span>;</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">v_right</span><span class="params">(n_right)</span></span>;</span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n_left; i++)</span><br><span class="line">    &#123;</span><br><span class="line">        v_left[i] = arr[start1 + i];</span><br><span class="line">    &#125; </span><br><span class="line">    <span class="keyword">for</span>(<span class="type">int</span> i= <span class="number">0</span>; i &lt; n_right; i++)</span><br><span class="line">    &#123;</span><br><span class="line">        v_right[i] = arr[start2 + i];</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="type">int</span> left = <span class="number">0</span>;</span><br><span class="line">    <span class="type">int</span> right = <span class="number">0</span>;</span><br><span class="line">    <span class="comment">// k 不能从0开始，因为分裂后的数不一定出现在0的位置，出现在数组的任意位置</span></span><br><span class="line">    <span class="type">int</span> k =start1;</span><br><span class="line">    <span class="comment">// 循环对比两边数组的元素，小的添加进数组；</span></span><br><span class="line">    <span class="comment">// 1. 当左、右两边的数组都还有数时；</span></span><br><span class="line">    <span class="keyword">while</span>(left &lt; n_left &amp;&amp; right &lt; n_right)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="comment">// 每一次判断</span></span><br><span class="line">        <span class="keyword">if</span>(v_left[left] &lt; v_right[right])</span><br><span class="line">        &#123;   </span><br><span class="line">            arr[k++] = v_left[left++];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">else</span></span><br><span class="line">        &#123;</span><br><span class="line">            arr[k++] = v_right[right++];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// 2. 当右边的数组已经空了，左边的数组还没空时；</span></span><br><span class="line">    <span class="keyword">while</span>(left &lt; n_left)</span><br><span class="line">    &#123;</span><br><span class="line">        arr[k++] = v_left[left++];</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// 3. 当左边的数组已经空了，右边的数组还没空时；</span></span><br><span class="line">    <span class="keyword">while</span>(right &lt; n_right)</span><br><span class="line">    &#123;</span><br><span class="line">        arr[k++] = v_right[right++];</span><br><span class="line">    &#125;</span><br><span class="line">&#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h4 id="6-快速排序"><a href="#6-快速排序" class="headerlink" title="6. 快速排序"></a>6. 快速排序</h4><ul>
<li><strong>思路</strong>：</li>
</ul>
<p>初始设置中轴数为最左边的元素，然后开始比，比他小的放在左边，大的放在右边，再更新中轴数位置；</p>
<p>继续左边数组：初始设置中轴数为最左边的元素，然后开始比，比他小的放在左边，大的放在右边，再更新中轴数位置；</p>
<p>继续右边数组：初始设置中轴数为最左边的元素，然后开始比，比他小的放在左边，大的放在右边，再更新中轴数位置；</p>
<p>直到数组分到最后只有一个元素的时候递归停止；</p>
<p>自顶向下的递归；</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<p>代码主要包括：</p>
<ol>
<li>需要一个递归函数：将一个数组分成左右两个数组，一直分到只剩下一个元素（递归的终止条件）；</li>
<li>需要一个找中轴数的函数：每次要分的时候需要找到中轴数才能分，因此还需要一个找中轴数的函数；</li>
<li>算法的核心就如何找中轴数（1.将中轴数初始值设为最左边的元素 2. 利用left、right两个下标，left找比它大的数，right找比它小的数，找到后就互换 3. 直到 left 下标超过 right，则更新中轴数）</li>
</ol>
<ul>
<li><strong>结果：</strong></li>
</ul>
<p>时间复杂度 O(nlogn) 	空间复杂度O(1) 	原地排序 	非稳定排序</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"><span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        <span class="type">int</span> n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">if</span>(n &lt; <span class="number">1</span>) <span class="keyword">return</span> nums;</span><br><span class="line"></span><br><span class="line">        <span class="built_in">QuickSort</span>(nums, <span class="number">0</span>, nums.<span class="built_in">size</span>()<span class="number">-1</span>);</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">QuickSort</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; arr, <span class="type">int</span> left, <span class="type">int</span> right)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(left &lt; right)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 中轴位置(每次都要重新获取中轴位置)</span></span><br><span class="line">            <span class="type">int</span> mid = <span class="built_in">sort</span>(arr, left, right);</span><br><span class="line">            <span class="comment">// 左边快速排序</span></span><br><span class="line">            <span class="built_in">QuickSort</span>(arr, left, mid<span class="number">-1</span>);</span><br><span class="line">            <span class="comment">// 右边快速排序</span></span><br><span class="line">            <span class="built_in">QuickSort</span>(arr, mid+<span class="number">1</span>, right);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">sort</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; arr, <span class="type">int</span> left, <span class="type">int</span> right)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 中轴值</span></span><br><span class="line">        <span class="type">int</span> mids = arr[left];</span><br><span class="line">        <span class="comment">// 左右下标</span></span><br><span class="line">        <span class="type">int</span> l = left +<span class="number">1</span>;</span><br><span class="line">        <span class="type">int</span> r = right;</span><br><span class="line">        <span class="comment">// 一个大循环，持续寻找</span></span><br><span class="line">        <span class="keyword">while</span>(<span class="literal">true</span>)&#123;</span><br><span class="line">            <span class="comment">// 1. 写一个循环，控制从左往右找，找到比中轴大的就停下；</span></span><br><span class="line">            <span class="keyword">while</span>(l&lt;=r &amp;&amp; arr[l] &lt;= mids) &#123;l++;&#125;</span><br><span class="line">            <span class="comment">// 2. 写一个循环，控制从右往左找，找到比中轴小的就停下；</span></span><br><span class="line">            <span class="keyword">while</span>(l&lt;=r &amp;&amp; arr[r] &gt;= mids) &#123;r--;&#125;</span><br><span class="line">            <span class="comment">// 3. 如果右 &lt; 左，则停止寻找</span></span><br><span class="line">            <span class="keyword">if</span>(l&gt;=r) <span class="keyword">break</span>;</span><br><span class="line">            <span class="comment">// 停下后互换，左右位置的值</span></span><br><span class="line">            <span class="type">int</span> temp = arr[l];</span><br><span class="line">            arr[l] = arr[r];</span><br><span class="line">            arr[r] = temp;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 找到中轴位置为右下标所处的位置，互换中轴</span></span><br><span class="line">        <span class="type">int</span> temp = arr[left];</span><br><span class="line">        arr[left] = arr[r];</span><br><span class="line">        arr[r] = temp;</span><br><span class="line">        <span class="comment">// 返回中轴位置</span></span><br><span class="line">        <span class="keyword">return</span> r;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h4 id="7-堆排序"><a href="#7-堆排序" class="headerlink" title="7. 堆排序"></a>7. 堆排序</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2.png"></p>
<ul>
<li>思路：</li>
</ul>
<p>通过建立大、小顶堆，依次交换顶堆元素和最后一个元素达到排序的目的；</p>
<ul>
<li>程序：</li>
</ul>
<ol>
<li><p>需要一个建堆函数：初始维护的点下标为：n&#x2F;2 - 1，再依次维护每一个下标即可</p>
</li>
<li><p>需要一个维护堆函数：判断维护点和其两个子节点哪个大，互换；然后判断下一个，继续互换；继续下一个…</p>
</li>
<li><p>需要一个堆排序函数：堆顶元素与最后一个元素交换、维护堆顶元素、删除最后一个元素；再次交换、维护、删除；….</p>
</li>
</ol>
<ul>
<li><strong>结果：</strong></li>
</ul>
<p>时间复杂度 O(nlogn) 	空间复杂度O(1) 	原地排序 	非稳定排序</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br><span class="line">71</span><br><span class="line">72</span><br><span class="line">73</span><br><span class="line">74</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"><span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        <span class="type">int</span> n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">if</span>(n &lt; <span class="number">1</span>) <span class="keyword">return</span> nums;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 建堆（大顶堆）</span></span><br><span class="line">        <span class="built_in">make</span>(nums, nums.<span class="built_in">size</span>());</span><br><span class="line">        <span class="comment">// 堆排序</span></span><br><span class="line">        <span class="built_in">Heap_Sort</span>(nums, nums.<span class="built_in">size</span>());</span><br><span class="line">        </span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// 建堆</span></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">make</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; arr, <span class="type">int</span> n)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 维护堆性质的第一个元素的下标：n/2 - 1</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = n/<span class="number">2</span> <span class="number">-1</span>; i &gt;=<span class="number">0</span>; i--)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 维护每一个堆元素</span></span><br><span class="line">            <span class="built_in">heapify</span>(arr, n, i);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// 堆排序</span></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">Heap_Sort</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp;arr, <span class="type">int</span> n)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 创建一个循环，来获取最后一个元素下标</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = n - <span class="number">1</span>; i &gt; <span class="number">0</span>; i--)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 交换堆顶元素和最后一个元素</span></span><br><span class="line">            <span class="built_in">swap</span>(arr[<span class="number">0</span>], arr[i]);</span><br><span class="line">            <span class="comment">// 维护堆顶元素</span></span><br><span class="line">            <span class="built_in">heapify</span>(arr, i, <span class="number">0</span>);</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// 维护堆</span></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">heapify</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; arr, <span class="type">int</span> n, <span class="type">int</span> i)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 父节点下标</span></span><br><span class="line">        <span class="type">int</span> parent = i;</span><br><span class="line">        <span class="comment">// 左子节点下标</span></span><br><span class="line">        <span class="type">int</span> lson = <span class="number">2</span> * i + <span class="number">1</span>;</span><br><span class="line">        <span class="comment">// 右子节点下标</span></span><br><span class="line">        <span class="type">int</span> rson = <span class="number">2</span> * i + <span class="number">2</span>;</span><br><span class="line">        <span class="comment">// 父节点与 左子节点比大小（前提是左子节点存在）</span></span><br><span class="line">        <span class="keyword">if</span>(lson &lt; n &amp;&amp; arr[parent] &lt; arr[lson])</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 互换元素</span></span><br><span class="line">            <span class="built_in">swap</span>(arr[parent], arr[lson]);</span><br><span class="line">            <span class="comment">// 接着维护互换后的子节点</span></span><br><span class="line">            <span class="built_in">heapify</span>(arr, n, lson);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 父节点与 右子节点比大小（前提是右子节点存在）</span></span><br><span class="line">        <span class="keyword">if</span>(rson &lt; n &amp;&amp; arr[parent] &lt; arr[rson])</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 互换元素</span></span><br><span class="line">            <span class="built_in">swap</span>(arr[parent], arr[rson]);</span><br><span class="line">            <span class="comment">// 接着维护互换后的子节点</span></span><br><span class="line">            <span class="built_in">heapify</span>(arr, n, rson);</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// 交换元素</span></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">swap</span><span class="params">(<span class="type">int</span>&amp; a, <span class="type">int</span>&amp; b)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="type">int</span> temp = a;</span><br><span class="line">        a = b;</span><br><span class="line">        b = temp;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    </span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h4 id="8-计数排序"><a href="#8-计数排序" class="headerlink" title="8. 计数排序"></a>8. 计数排序</h4><ul>
<li><strong>思路</strong></li>
</ul>
<p>之前的算法都是通过比较大小来排序，这个算法是通过创建一个统计每个元素出现次数的数组，该数组的尺寸为数组元素最大值+1个，统计完之后再恢复到原数组中</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>找出数组中的最大值</li>
<li>创建一个初始值为0，大小为 max + 1的数组</li>
<li>统计元素出现的次数</li>
<li>把统计的数据汇总回原数组中</li>
</ol>
<ul>
<li><strong>结果</strong></li>
</ul>
<p>时间复杂度 O(n+k) 	空间复杂度O(k) 	非原地排序 	稳定排序</p>
<p>k 表示临时数组的大小</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"><span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        <span class="type">int</span> n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">if</span>(n &lt; <span class="number">2</span>) <span class="keyword">return</span> nums;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 找到数组的最大值</span></span><br><span class="line">        <span class="type">int</span> max = nums[<span class="number">0</span>];</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">1</span>; i &lt; n; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span>(nums[i] &gt; max)&#123;max = nums[i];&#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 创建大小为max的临时数组</span></span><br><span class="line">        <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">temp</span><span class="params">(max+<span class="number">1</span>, <span class="number">0</span>)</span></span>;</span><br><span class="line">        <span class="comment">// 统计元素出现的次数</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i=<span class="number">0</span>; i&lt; n; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 某个元素出现次数自增1</span></span><br><span class="line">            temp[nums[i]]++;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="type">int</span> s =<span class="number">0</span>;</span><br><span class="line">        <span class="comment">// 把数组统计好的数据汇总到原数组中</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i=<span class="number">0</span>; i&lt;=max; i++) <span class="comment">// 遍历每一个容器</span></span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = temp[i]; j&gt;<span class="number">0</span>; j--) <span class="comment">// 容器的值出现几次</span></span><br><span class="line">            &#123;</span><br><span class="line">                nums[s++] = i;</span><br><span class="line">            &#125;     </span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="9-桶排序"><a href="#9-桶排序" class="headerlink" title="9. 桶排序"></a>9. 桶排序</h4><ul>
<li><strong>思路</strong></li>
</ul>
<p>确定桶的数目可以直接设定，或者根据最大值和最小值来定；根据数组中的数值的大小来确定用什么方式来分，比如0-99，就用nums[i]&#x2F;10 来分；然后分别放入桶中，对每个桶内的数排序；再合并桶的数；</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>找出最大最小值，确定桶的数目，大小</li>
<li>分别装入每个桶中（用vector 的二维数组）</li>
<li>对每个桶进行排序（冒泡排序）</li>
<li>合并</li>
</ol>
<ul>
<li><strong>结果</strong></li>
</ul>
<p>时间复杂度 O(n+k) 	空间复杂度O(n+k) 	非原地排序 	稳定排序</p>
<p>k 表示桶的个数</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"><span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">sortArray</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        <span class="type">int</span> n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">if</span>(n &lt; <span class="number">2</span>) <span class="keyword">return</span> nums;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 找到数组的最大值和最小值</span></span><br><span class="line">        <span class="type">int</span> max = nums[<span class="number">0</span>];</span><br><span class="line">        <span class="type">int</span> min = nums[<span class="number">0</span>];</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">1</span>; i &lt; n; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span>(nums[i] &gt; max)&#123;max = nums[i];&#125;</span><br><span class="line">            <span class="keyword">if</span>(nums[i] &lt; min)&#123;min = nums[i];&#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 设置桶的大小为10</span></span><br><span class="line">        <span class="type">const</span> <span class="type">int</span> bucketNum = max + <span class="number">1</span>;</span><br><span class="line">        <span class="comment">// 创建一个二维数组</span></span><br><span class="line">        vector&lt;vector&lt;<span class="type">int</span>&gt;&gt; <span class="built_in">bucket</span>(bucketNum);</span><br><span class="line">        <span class="comment">// 将每一个元素放入对应的桶中</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            bucket[nums[i]].<span class="built_in">push_back</span>(nums[i]);</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 对每个桶进行排序</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; bucketNum; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 冒泡排序</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> k = <span class="number">0</span>; k &lt; bucket[i].<span class="built_in">size</span>(); k++)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">1</span>; j&lt; bucket[i].<span class="built_in">size</span>(); j++)</span><br><span class="line">                &#123;</span><br><span class="line">                    <span class="comment">// 如果 左&gt; 右</span></span><br><span class="line">                    <span class="keyword">if</span>(bucket[i][j] &lt; bucket[i][j<span class="number">-1</span>])&#123;</span><br><span class="line">                        <span class="comment">// 交换元素</span></span><br><span class="line">                        <span class="type">int</span> temp = bucket[i][j];</span><br><span class="line">                        bucket[i][j] = bucket[i][j<span class="number">-1</span>];</span><br><span class="line">                        bucket[i][j<span class="number">-1</span>] = temp;</span><br><span class="line">                    &#125;</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="type">int</span> k = <span class="number">0</span>;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 放回原数组</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; bucketNum; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">0</span>; j &lt; bucket[i].<span class="built_in">size</span>(); j++)</span><br><span class="line">            &#123;</span><br><span class="line">                nums[k++] = bucket[i][j];</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> nums;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="10-基数排序"><a href="#10-基数排序" class="headerlink" title="10. 基数排序"></a>10. 基数排序</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>比较个位、十位、百位</li>
</ol>
<ul>
<li>非优化版本</li>
<li>优化版本</li>
</ul>
<p>栈 堆 红黑树 </p>
<h1 id="面试题"><a href="#面试题" class="headerlink" title="面试题"></a>面试题</h1><h3 id="数组（6）"><a href="#数组（6）" class="headerlink" title="数组（6）"></a>数组（6）</h3><h4 id="1-两数之和"><a href="#1-两数之和" class="headerlink" title="1. 两数之和"></a>1. 两数之和</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>target - 遍历到的数 &#x3D; 我们要找的数；</li>
<li>建立哈希表，在哈希表中寻找我想要的数，若找到，则直接返回 value，若没找到，添加进表中；这样就避免的重复计算；</li>
<li>就算第一次错过，第一次就会把这个数添加进表中，第二次一定会搜索到这个数；所有数都会遍历一边，有的话一定会被找到；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> </span><br><span class="line">&#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">twoSum</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums, <span class="type">int</span> target)</span> </span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 创建一个哈希map，用来存放已经遍历过的元素（避免重复）</span></span><br><span class="line">        std::unordered_map&lt;<span class="type">int</span>, <span class="type">int</span>&gt; maps;</span><br><span class="line">        <span class="comment">// 遍历每一个元素</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 计算需要查找的元素 </span></span><br><span class="line">            <span class="type">int</span> t = target - nums[i];</span><br><span class="line">            <span class="comment">// 检查哈希表中是否包含元素</span></span><br><span class="line">            <span class="keyword">auto</span> it = maps.<span class="built_in">find</span>(t);</span><br><span class="line">            <span class="comment">// 若有， 则输出当前遍历元素和哈希表的下标</span></span><br><span class="line">            <span class="keyword">if</span>(it != maps.<span class="built_in">end</span>())</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">return</span> &#123;it-&gt;second, i&#125;;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 若没有，则将元素加入到哈希表中</span></span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">            &#123;</span><br><span class="line">                maps.<span class="built_in">insert</span>(std::<span class="built_in">pair</span>&lt;<span class="type">int</span>, <span class="type">int</span>&gt;(nums[i], i));</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> &#123;&#125;;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="70-爬楼梯"><a href="#70-爬楼梯" class="headerlink" title="70. 爬楼梯"></a>70. 爬楼梯</h4><ul>
<li><strong>程序</strong></li>
</ul>
<p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/84_%5B9B8E%5D5_YXXDPEE0AY@9.png"></p>
<ol>
<li><strong>递归法：</strong>把爬楼梯看成是<strong>下楼梯</strong>，假设有6层台阶，可以把下楼梯的方式看作是一个二叉树，每次走只有两种可能；</li>
<li>每次下楼梯的选择 f(n)  &#x3D; 走一步之后的结果 f(n-1) + 走两步之后的结果 f(n-2)；（<strong>递归函数</strong>）</li>
<li><strong>哈希表法：</strong>每次在哈希表寻找之前有走过的路，若找到，则直接返回，避免重复计算；若没找到，则把该条路径存入表中</li>
<li>终止条件：直到只剩下一个或两个台阶；（<strong>递归终止条件</strong>）</li>
</ol>
<ul>
<li>纯粹递归（会出现大量重复计算，时间复杂度 O(n^2)）</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    std::unordered_map&lt;<span class="type">int</span>, <span class="type">int</span>&gt; maps;</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">climbStairs</span><span class="params">(<span class="type">int</span> n)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 递归的终止条件：如果最后只剩下一个或两个台阶，直接返回1，2</span></span><br><span class="line">        <span class="keyword">if</span>(n == <span class="number">1</span>) <span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span>(n == <span class="number">2</span>) <span class="keyword">return</span> <span class="number">2</span>;</span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">climbStairs</span>(n<span class="number">-1</span>) + <span class="built_in">climbStairs</span>(n<span class="number">-2</span>);</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>带哈希表的解法✔（时间复杂度 O(n) ）</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    std::unordered_map&lt;<span class="type">int</span>, <span class="type">int</span>&gt; maps;</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">climbStairs</span><span class="params">(<span class="type">int</span> n)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 递归的终止条件：如果最后只剩下一个或两个台阶，直接返回1，2</span></span><br><span class="line">        <span class="keyword">if</span>(n == <span class="number">1</span>) <span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span>(n == <span class="number">2</span>) <span class="keyword">return</span> <span class="number">2</span>;</span><br><span class="line">        <span class="comment">// 哈希表迭代器指针</span></span><br><span class="line">        unordered_map&lt;<span class="type">int</span>, <span class="type">int</span>&gt;::iterator iter = maps.<span class="built_in">find</span>(n);</span><br><span class="line">        <span class="comment">// 在哈希表中找元素</span></span><br><span class="line">        <span class="keyword">if</span>(iter != maps.<span class="built_in">end</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 如果找到，则返回value</span></span><br><span class="line">            <span class="keyword">return</span> iter-&gt;second;     </span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">else</span></span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 如果没找到，就存进去，并返回本次结果</span></span><br><span class="line">            maps[n] = <span class="built_in">climbStairs</span>(n<span class="number">-1</span>) + <span class="built_in">climbStairs</span>(n<span class="number">-2</span>);</span><br><span class="line">            <span class="keyword">return</span> <span class="built_in">climbStairs</span>(n<span class="number">-1</span>) + <span class="built_in">climbStairs</span>(n<span class="number">-2</span>);</span><br><span class="line">        &#125;       </span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="88-合并两个数组"><a href="#88-合并两个数组" class="headerlink" title="88. 合并两个数组"></a>88. 合并两个数组</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>双头指针正序：</strong>合并两个升序数组，用双指针方法；新建一个临时数组，比较两个头指针的数，小的存进临时数组，然后指针右移；当某个数组遍历结束，直接取另一个数组的元素放进临时数组即可</li>
<li><strong>双尾指针倒序：</strong>创建三个指针，两个指向左右数组的尾指针 s1、s2，一个指向左数组的尾指针s3 ；然后比较左右两个尾指针s2、s3的数，大的存进做尾指针 s3 指向的元素，然后大的指针 s2 左移，尾指针 s3 也左移</li>
<li>注意数组的有效元素的个数！！！</li>
</ol>
<ul>
<li>双头指针正序（时间复杂度 O(m+n) ）</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="type">void</span> <span class="title">merge</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums1, <span class="type">int</span> m, vector&lt;<span class="type">int</span>&gt;&amp; nums2, <span class="type">int</span> n)</span> </span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">    <span class="comment">// 利用上两个数组都是非递减序列的条件，双指针</span></span><br><span class="line">    <span class="comment">// 1. 定义两个指针，分别指向数组的头，一个临时数组v，大小为m+n</span></span><br><span class="line">    <span class="type">int</span> left = <span class="number">0</span>;</span><br><span class="line">    <span class="type">int</span> right = <span class="number">0</span>;</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">v</span><span class="params">(m+n)</span></span>;</span><br><span class="line">    <span class="type">int</span> k = <span class="number">0</span>;</span><br><span class="line">    <span class="comment">// 2. 比较两个数，小的添加进数组，指针+1；</span></span><br><span class="line">    <span class="comment">// 两个数组都未取完</span></span><br><span class="line">    <span class="keyword">while</span>(left &lt; m &amp;&amp; right &lt; n)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="keyword">if</span>(nums1[left] &lt;= nums2[right])</span><br><span class="line">        &#123;</span><br><span class="line">            v[k++] = nums1[left++];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">else</span></span><br><span class="line">        &#123;</span><br><span class="line">            v[k++] = nums2[right++];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// nums2 取完了</span></span><br><span class="line">    <span class="keyword">while</span>(left &lt; m)</span><br><span class="line">    &#123;</span><br><span class="line">        v[k++] = nums1[left++];</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// nums1 取完了</span></span><br><span class="line">    <span class="keyword">while</span>(right &lt; n)</span><br><span class="line">    &#123;</span><br><span class="line">        v[k++] = nums2[right++];</span><br><span class="line">    &#125;</span><br><span class="line">    nums1 = v;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>双尾指针倒序（时间复杂度 O(m+n) ）</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">merge</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums1, <span class="type">int</span> m, vector&lt;<span class="type">int</span>&gt;&amp; nums2, <span class="type">int</span> n)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        <span class="comment">// 1. 定义双指针，分别指向两个数组的尾部</span></span><br><span class="line">        <span class="type">int</span> left = m<span class="number">-1</span>;</span><br><span class="line">        <span class="type">int</span> right = n<span class="number">-1</span>;</span><br><span class="line">        <span class="type">int</span> k = m + n - <span class="number">1</span>;</span><br><span class="line">        <span class="comment">// 2. 判断两个数大小，大的赋值给nums1，指针左移动</span></span><br><span class="line">        <span class="keyword">while</span>(right &gt;= <span class="number">0</span> &amp;&amp; left &gt;=<span class="number">0</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span>(nums1[left] &lt; nums2[right])</span><br><span class="line">            &#123;</span><br><span class="line">                nums1[k--] = nums2[right--];</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">            &#123;</span><br><span class="line">                nums1[k--] = nums1[left--];</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">while</span> (right&gt;=<span class="number">0</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            nums1[k--] = nums2[right--];</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="283-移动零"><a href="#283-移动零" class="headerlink" title="283. 移动零"></a>283. 移动零</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><p><strong>相邻快慢双指针法</strong>：快指针用于找非零元素，慢指针用于记录非零元素</p>
<p>step1. 创建两个指针 left（指向非0 元素） 、right（遍历所有元素）</p>
<p>step2. 用right 遍历数组，如果 right 指向元素不等于0，则将 left 和right 的值交换，left 右移动准备接收下一个非零元素</p>
</li>
</ol>
<ul>
<li><strong>结果</strong></li>
</ul>
<p>时间复杂度 O(n) 	空间复杂度O(1) </p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">moveZeroes</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 定义左右指针，左指针放非零元素，右指针放0</span></span><br><span class="line">        <span class="type">int</span> left = <span class="number">0</span>;</span><br><span class="line">        <span class="type">int</span> right = <span class="number">0</span>;</span><br><span class="line">        <span class="comment">// 遍历数组</span></span><br><span class="line">        <span class="keyword">for</span>(; right &lt; nums.<span class="built_in">size</span>();right++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 若不等于0，则左右指针的值交换，左指针往前移动，右指针往前移动</span></span><br><span class="line">            <span class="keyword">if</span>(nums[right] != <span class="number">0</span>)</span><br><span class="line">            &#123;   </span><br><span class="line">                <span class="comment">// 交换数值</span></span><br><span class="line">                <span class="built_in">swap</span>(nums[left], nums[right]);</span><br><span class="line">                left++;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 若为0，则移动右指针</span></span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">// 交换函数</span></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">swap</span><span class="params">(<span class="type">int</span> &amp;a, <span class="type">int</span> &amp; b)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="type">int</span> temp = a;</span><br><span class="line">        a = b;</span><br><span class="line">        b = temp;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="448-找到所有数组中消失的数字"><a href="#448-找到所有数组中消失的数字" class="headerlink" title="448. 找到所有数组中消失的数字"></a>448. 找到所有数组中消失的数字</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><p><strong>利用数组元素与下标索引的特殊关系:<strong>因为数组的范围是 [1, n]，</strong>自然可以找到数组元素和下标索引的关系</strong>；数组元素 &#x3D; 下标索引+1</p>
</li>
<li><p>第一次遍历数组，如果数组元素是正，就把下标为（数组元素-1）的元素标记一下（取负数）；如果不是正数，就还原为正数后再把下标为（数组元素-1）的元素标记为负数；</p>
</li>
<li><p>第二次遍历数组，如果不是负数，说明该下标就是消失的数字。</p>
</li>
<li><p><strong>哈希表法：</strong>把数组元素存入哈希表（自动删除重复元素），再遍历 [1, n]个数，每次查找哈希表，查不到的说明这就是消失的数字；</p>
</li>
</ol>
<ul>
<li><strong>利用数组元素与下标索引的特殊关系</strong>（空间复杂度 O(1)）</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">findDisappearedNumbers</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i&lt; nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 还原数组元素取绝对值(因为有可能原本的数值被取负了)</span></span><br><span class="line">            <span class="type">int</span> l = <span class="built_in">abs</span>(nums[i]);</span><br><span class="line">            <span class="keyword">if</span>(nums[l - <span class="number">1</span>] &gt; <span class="number">0</span>) </span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 取负</span></span><br><span class="line">                nums[l - <span class="number">1</span>] *= <span class="number">-1</span>;</span><br><span class="line">            &#125;</span><br><span class="line"></span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 遍历所有元素，不是负数的数组元素的下标+1存入新数组中</span></span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; v;</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span>(nums[i] &gt; <span class="number">0</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                v.<span class="built_in">push_back</span>(i+<span class="number">1</span>);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> v;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">abs</span><span class="params">(<span class="type">int</span> n)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">return</span> n = n &gt; <span class="number">0</span>? n: -n;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>哈希表法（空间复杂度 O(n)）</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">findDisappearedNumbers</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 缺失值数组</span></span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; miss_value;</span><br><span class="line">        <span class="comment">// 创建哈希表</span></span><br><span class="line">        std::unordered_map&lt;<span class="type">int</span>, <span class="type">int</span>&gt; Hash;</span><br><span class="line">        <span class="comment">// 将数组放入哈希表中</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> x : nums)</span><br><span class="line">        &#123;</span><br><span class="line">            Hash.<span class="built_in">insert</span>(std::<span class="built_in">pair</span>&lt;<span class="type">int</span>, <span class="type">int</span>&gt;(x, <span class="number">0</span>));</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 遍历数字1-n,若找到则循环，</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i=<span class="number">1</span>; i &lt;= nums.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">auto</span> it = Hash.<span class="built_in">find</span>(i);</span><br><span class="line">            <span class="comment">// 若没找到，则保持到缺失值数组中</span></span><br><span class="line">            <span class="keyword">if</span>(it == Hash.<span class="built_in">end</span>())</span><br><span class="line">            &#123;</span><br><span class="line">                miss_value.<span class="built_in">push_back</span>(i);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> miss_value;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="4-寻找两个正序数组的中位数"><a href="#4-寻找两个正序数组的中位数" class="headerlink" title="4. 寻找两个正序数组的中位数"></a>4. 寻找两个正序数组的中位数</h4><ul>
<li>程序</li>
</ul>
<ol>
<li>先将两个数组排序，合并成一个数组</li>
<li>若数组个数为奇数 n ：则中位数为 ans[n &#x2F; 2];</li>
<li>若数组个数为偶数 n：则中位数为均值 (ans[n &#x2F; 2] + ans[n &#x2F; 2 - 1]) &#x2F; 2.0</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">double</span> <span class="title">findMedianSortedArrays</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums1, vector&lt;<span class="type">int</span>&gt;&amp; nums2)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 数组长度</span></span><br><span class="line">        <span class="type">int</span> n = nums1.<span class="built_in">size</span>(), m = nums2.<span class="built_in">size</span>();</span><br><span class="line">        <span class="comment">// 判空</span></span><br><span class="line">        <span class="keyword">if</span>(n == <span class="number">0</span> &amp;&amp; m == <span class="number">0</span>) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        <span class="comment">// 创建一个数组用来存放排序后的</span></span><br><span class="line">        <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">ans</span><span class="params">(n + m)</span></span>;</span><br><span class="line">        <span class="comment">// 初始化指针</span></span><br><span class="line">        <span class="type">int</span> i = <span class="number">0</span>, j = <span class="number">0</span>, k = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">while</span>(i &lt;  n &amp;&amp; j &lt; m)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span>(nums1[i] &lt; nums2[j]) ans[k++] = nums1[i++];</span><br><span class="line">            <span class="keyword">else</span> ans[k++] = nums2[j++];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">while</span>(i &lt; n)</span><br><span class="line">        &#123;</span><br><span class="line">            ans[k++] = nums1[i++];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">while</span>(j &lt; m)</span><br><span class="line">        &#123;</span><br><span class="line">            ans[k++] = nums2[j++];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 如果是奇数</span></span><br><span class="line">        <span class="keyword">if</span>((n + m) % <span class="number">2</span> != <span class="number">0</span>) <span class="keyword">return</span> ans[(n + m)/<span class="number">2</span>];</span><br><span class="line">        <span class="comment">// 如果是偶数</span></span><br><span class="line">        <span class="keyword">else</span> <span class="built_in">return</span> (ans[(n + m) / <span class="number">2</span>] + ans[(n + m) / <span class="number">2</span> - <span class="number">1</span>]) / <span class="number">2.0</span>; </span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h3 id="链表（14）"><a href="#链表（14）" class="headerlink" title="链表（14）"></a>链表（14）</h3><p>走到第 n 个节点</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 定义虚拟节点</span></span><br><span class="line"><span class="keyword">auto</span> dummy = <span class="keyword">new</span> <span class="built_in">ListNode</span>(<span class="number">-1</span>);</span><br><span class="line">dummy-&gt;next = head;</span><br><span class="line"><span class="comment">// 定义第 n 个节点</span></span><br><span class="line"><span class="keyword">auto</span> n = dummy;</span><br><span class="line"><span class="comment">// 走到第 n 个节点</span></span><br><span class="line"><span class="keyword">while</span>(n--) n = n-&gt;next;</span><br></pre></td></tr></table></figure>

<h4 id="21-合并两个有序链表"><a href="#21-合并两个有序链表" class="headerlink" title="21. 合并两个有序链表"></a>21. 合并两个有序链表</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>双头指针正序：</strong>跟合并两个有序数组一样的做法，定义两个头指针，比较头数组元素的大小，小的存入临时链表，并指针右移动；移动结束后，直接把剩余的链表拼接即可；</li>
<li>创建一个头链表，用于返回值；再创建一个临时链表，用于拼接</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">mergeTwoLists</span><span class="params">(ListNode* list1, ListNode* list2)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 边界条件</span></span><br><span class="line">        <span class="keyword">if</span>(list1 == <span class="literal">NULL</span>) <span class="keyword">return</span> list2;</span><br><span class="line">        <span class="keyword">if</span>(list2 == <span class="literal">NULL</span>) <span class="keyword">return</span> list1;</span><br><span class="line">        <span class="comment">// 创建一个临时节点，当指向这个节点时可用于返回</span></span><br><span class="line">        ListNode *head = <span class="keyword">new</span> ListNode;</span><br><span class="line">        <span class="comment">// 再创建一个新链表，用于存放</span></span><br><span class="line">        ListNode *p = head;</span><br><span class="line">        <span class="comment">// 两个链表都不空时</span></span><br><span class="line">        <span class="keyword">while</span> (list1 != <span class="literal">NULL</span> &amp;&amp; list2 != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 判断哪个小，小的保存，指针右移动</span></span><br><span class="line">            <span class="keyword">if</span>(list1-&gt;val &lt; list2-&gt;val)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 连接list节点</span></span><br><span class="line">                p-&gt;next = list1;</span><br><span class="line">                <span class="comment">// 指向下一个节点</span></span><br><span class="line">                list1 = list1-&gt;next;</span><br><span class="line">                <span class="comment">// 移动到该节点</span></span><br><span class="line">                p = p-&gt;next;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 连接list节点</span></span><br><span class="line">                p-&gt;next = list2;</span><br><span class="line">                <span class="comment">// 指向下一个节点</span></span><br><span class="line">                list2 = list2-&gt;next;</span><br><span class="line">                <span class="comment">// 移动到该节点</span></span><br><span class="line">                p = p-&gt;next;</span><br><span class="line">            &#125;</span><br><span class="line"></span><br><span class="line">        &#125;</span><br><span class="line">       <span class="keyword">if</span>(list1 != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            p-&gt;next = list1;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span>(list2 != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            p-&gt;next = list2;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> head-&gt;next;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="83-删除排序链表中的重复元素"><a href="#83-删除排序链表中的重复元素" class="headerlink" title="83. 删除排序链表中的重复元素"></a>83. 删除排序链表中的重复元素</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>判断相邻节点是否相同：</strong>遍历节点，判断当前节点和下一节点的值是否相同；若相同，则直接越过下一节点；若不相同，则指向下一个节点；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">deleteDuplicates</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御措施</span></span><br><span class="line">        <span class="keyword">if</span>(head == <span class="literal">NULL</span>) <span class="keyword">return</span> head;</span><br><span class="line">        <span class="comment">// 创建一个节点</span></span><br><span class="line">        ListNode *cur_node = head;</span><br><span class="line">        <span class="comment">// 首先要判断下一节点是否为空,只有不为空才能继续往下操作</span></span><br><span class="line">        <span class="keyword">while</span>(cur_node-&gt;next != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 判断当前节点的值和下一节点的值是否相等</span></span><br><span class="line">            <span class="keyword">if</span>(cur_node-&gt;val == cur_node-&gt;next-&gt;val)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 直接越过下一个节点，移动到下下一个</span></span><br><span class="line">                cur_node-&gt;next = cur_node-&gt;next-&gt;next;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 移动到下个节点</span></span><br><span class="line">                cur_node = cur_node-&gt;next;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> head;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="141-环形链表"><a href="#141-环形链表" class="headerlink" title="141. 环形链表"></a>141. 环形链表</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>相遇快慢双指针法：</strong>快指针一次走两步，慢指针一次走一步；有快有慢，是环形的话最终一定会相遇；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">hasCycle</span><span class="params">(ListNode *head)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御编码</span></span><br><span class="line">        <span class="keyword">if</span>(head == <span class="literal">NULL</span>) <span class="keyword">return</span> head;</span><br><span class="line">        <span class="comment">// 创建快慢指针</span></span><br><span class="line">        ListNode *fast_node = head;</span><br><span class="line">        ListNode *slow_node = head;</span><br><span class="line">        <span class="comment">// 先判断下一个节点和下下个节点是否存在，存在才能走</span></span><br><span class="line">        <span class="keyword">while</span>(fast_node-&gt;next != <span class="literal">NULL</span> &amp;&amp; fast_node-&gt;next-&gt;next != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 先走</span></span><br><span class="line">            fast_node = fast_node-&gt;next-&gt;next;</span><br><span class="line">            slow_node = slow_node-&gt;next;</span><br><span class="line">            <span class="comment">// 再判断是否相遇</span></span><br><span class="line">            <span class="keyword">if</span>(fast_node == slow_node) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="142-环形链表Ⅱ"><a href="#142-环形链表Ⅱ" class="headerlink" title="142. 环形链表Ⅱ"></a>142. 环形链表Ⅱ</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>相遇快慢双指针法：</strong>快指针一次走两步，慢指针一次走一步；有快有慢，是环形的话最终一定会相遇；相遇后记录</li>
<li>相遇后将<strong>慢指针</strong>重新指向头部，同步速度；继续走，再次相遇就是环的头</li>
<li>创建两个快慢指针；创建一个环形标志位</li>
<li>循环：快指针一次走两步，慢指针一次走一步；每次判断快慢指针是否相等，相等就将标志位置为 true，并退出循环；不相遇的话就继续循环，直到 head-&gt;next &#x3D; NULL || head-&gt;next-&gt;next &#x3D;&#x3D; NULL，最终返回 NULL;</li>
<li>如果有环，则将快指针和慢指针同步速度；一直判断是否相等，再次相等就是环的头；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br></pre></td><td class="code"><pre><span class="line"></span><br><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode *<span class="title">detectCycle</span><span class="params">(ListNode *head)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御编码</span></span><br><span class="line">        <span class="keyword">if</span>(head == <span class="literal">NULL</span>) <span class="keyword">return</span> head;</span><br><span class="line">        <span class="comment">// 定义快慢指针</span></span><br><span class="line">        ListNode *fast_node = head;</span><br><span class="line">        ListNode *slow_node = head;</span><br><span class="line">        <span class="comment">// 回环标志位</span></span><br><span class="line">        <span class="type">bool</span> flag = <span class="literal">false</span>;</span><br><span class="line">        <span class="comment">// 首先判断是否能相遇</span></span><br><span class="line">        <span class="comment">// 1. 判断下个节点和下下个节点是否存在    </span></span><br><span class="line">        <span class="keyword">while</span> (fast_node-&gt;next != <span class="literal">NULL</span> &amp;&amp; fast_node-&gt;next-&gt;next != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 开始走</span></span><br><span class="line">            fast_node = fast_node-&gt;next-&gt;next;</span><br><span class="line">            slow_node = slow_node-&gt;next;</span><br><span class="line">            <span class="comment">// 判断是否相遇</span></span><br><span class="line">            <span class="keyword">if</span>(fast_node == slow_node)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 若相遇，则回环标志位置1</span></span><br><span class="line">                flag = <span class="literal">true</span>;</span><br><span class="line">                <span class="comment">// 退出循环</span></span><br><span class="line">                <span class="keyword">break</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 2. 如果回环存在</span></span><br><span class="line">        <span class="keyword">if</span>(flag == <span class="literal">true</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 3. 慢节点指向头，并且快慢节点同步速度</span></span><br><span class="line">            slow_node = head;</span><br><span class="line">            <span class="comment">// 4. 再次相遇一定是回环的头节点,循环判断快慢节点是否相同</span></span><br><span class="line">            <span class="keyword">while</span>(fast_node != slow_node)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 走</span></span><br><span class="line">                fast_node = fast_node-&gt;next;</span><br><span class="line">                slow_node = slow_node-&gt;next;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="keyword">return</span> slow_node;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">else</span></span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">return</span> <span class="literal">NULL</span>;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="160-相交链表"><a href="#160-相交链表" class="headerlink" title="160. 相交链表"></a>160. 相交链表</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.28(8.58).png"></p>
<ul>
<li><strong>思路</strong></li>
</ul>
<p>运动速度相同的两个物体，只要运动的总距离相同，那么他们最终一定会相遇；如果距离不同，那么不可能相遇；</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>交叉环双指针法：</strong>定义两个指针，指向两个节点头；每次往右移动，如果A节点移动到NULL，则再从B 节点头开始移动；同理，如果B 节点移动到NULL，则再从A节点开始移动；直到他们相交</li>
<li><strong>暴力穷举法：</strong>将 A 链表中的每一个节点都与 B 链表的所有节点对比；</li>
<li><strong>哈希表法：</strong>将 A 链表的节点存入哈希表中，遍历 B 链表的节点，每次判断是否出现再哈希表中，若出现，则返回结果；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode *<span class="title">getIntersectionNode</span><span class="params">(ListNode *headA, ListNode *headB)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 定义两个指针</span></span><br><span class="line">        ListNode *A = headA;</span><br><span class="line">        ListNode *B = headB;</span><br><span class="line">        <span class="comment">// 如果A节点和B节点不相等</span></span><br><span class="line">        <span class="keyword">while</span>(A != B)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 判断是否为null，是则交叉从头开始；不是则指向下一个节点</span></span><br><span class="line">            A = A == <span class="literal">NULL</span>? headB : A-&gt;next;</span><br><span class="line">            B = B == <span class="literal">NULL</span>? headA : B-&gt;next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> A;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br><span class="line"></span><br></pre></td></tr></table></figure>

<h4 id="206-反转链表"><a href="#206-反转链表" class="headerlink" title="206. 反转链表"></a>206. 反转链表</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/55.png"></p>
<ul>
<li>程序</li>
</ul>
<ol>
<li><strong>迭代法：</strong>遍历节点，使得每一个节点指向它的前一个节点；</li>
<li>断开节点1到节点2的连接，将节点1的next指向null；这样会导致节点2找不到了，因此需要创建一个临时变量来保存节点2的位置；一次循环；</li>
<li>迭代是从上处理完，再往下处理，再往下处理，从顶往下遍历操作；</li>
<li><strong>递归法：</strong>把问题拆分成若干个小问题，只要把小问题解决；再解决扩展到所有问题的时候的特殊处理；</li>
<li>递归是递归到叶节点，然后再从底往上遍历操作；</li>
</ol>
<ul>
<li>结果</li>
</ul>
<p>时间复杂度 O(n) 	空间复杂度 O(1) 	</p>
<p>只对链表遍历了一边，所以时间是 n</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">reverseList</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 上一个节点</span></span><br><span class="line">        ListNode *pre = <span class="literal">NULL</span>;</span><br><span class="line">        <span class="comment">// 当前节点</span></span><br><span class="line">        ListNode *cur = head;</span><br><span class="line">        <span class="comment">// 遍历节点，使得每一次当前节点指向上一个节点</span></span><br><span class="line">        <span class="keyword">while</span>(cur != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 记录下一个节点</span></span><br><span class="line">            ListNode *next = cur-&gt;next;</span><br><span class="line">            <span class="comment">// 指向上一个节点</span></span><br><span class="line">            cur-&gt;next = pre;</span><br><span class="line">            <span class="comment">// 更新上一个节点</span></span><br><span class="line">            pre = cur;</span><br><span class="line">            <span class="comment">// 更新当前节点</span></span><br><span class="line">            cur = next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> pre;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>递归</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">reverseList</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="comment">// 先递归到最后一个节点，然后再做处理；因此递归函数放在处理函数前；</span></span><br><span class="line">        <span class="keyword">if</span>(head == <span class="literal">NULL</span> || head-&gt;next == <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">return</span> head;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 保存链表头</span></span><br><span class="line">        ListNode* new_head = <span class="built_in">reverseList</span>(head-&gt;next);</span><br><span class="line">        <span class="comment">// 若当前head在节点4，则将节点5的next指向节点四机head</span></span><br><span class="line">        head-&gt;next-&gt;next = head;</span><br><span class="line">        <span class="comment">// 将节点4的next指向NULL，断开与节点5的连接，然后往上一级；</span></span><br><span class="line">        head-&gt;next = <span class="literal">NULL</span>;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">return</span> new_head;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="92-反转链表-Ⅱ"><a href="#92-反转链表-Ⅱ" class="headerlink" title="92. 反转链表 Ⅱ"></a>92. 反转链表 Ⅱ</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>找到 m-1, n  节点的位置；</li>
<li>保存 m , n+1 节点；（后面用于连接反转后的链表）</li>
<li>将 m —- n 的节点反转；</li>
<li>m-1-&gt;next 指向 n 节点，m-&gt; next 指向 n+1节点</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">reverseBetween</span><span class="params">(ListNode* head, <span class="type">int</span> m, <span class="type">int</span> n)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">if</span>(m == n) <span class="keyword">return</span> head;</span><br><span class="line">        <span class="comment">// 创建虚拟头节点</span></span><br><span class="line">        <span class="keyword">auto</span> dummy = <span class="keyword">new</span> <span class="built_in">ListNode</span>(<span class="number">-1</span>);</span><br><span class="line">        dummy-&gt;next = head;</span><br><span class="line">        <span class="comment">// 定义m-1 和 n 位置节点</span></span><br><span class="line">        <span class="keyword">auto</span> m_1 = dummy, nn = dummy;</span><br><span class="line">        <span class="comment">// 先找到 m-1 和 n 节点位置</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; m<span class="number">-1</span>; i++) m_1 = m_1-&gt;next;</span><br><span class="line">        <span class="keyword">while</span>(n--) nn = nn-&gt;next;</span><br><span class="line">        <span class="comment">// 保存 m 节点</span></span><br><span class="line">        <span class="keyword">auto</span> mm = m_1-&gt;next;</span><br><span class="line">        <span class="comment">// 保存 n+1 节点</span></span><br><span class="line">        <span class="keyword">auto</span> n_1 = nn-&gt;next;</span><br><span class="line">        <span class="comment">// 反转 m - n  链表</span></span><br><span class="line">        ListNode* pre;</span><br><span class="line">        ListNode* cur;</span><br><span class="line">        pre = <span class="literal">NULL</span>;</span><br><span class="line">        cur = mm;</span><br><span class="line">        <span class="keyword">while</span>(cur != n_1)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">auto</span> next = cur-&gt;next;</span><br><span class="line">            cur-&gt;next = pre;</span><br><span class="line">            pre = cur;</span><br><span class="line">            cur = next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 将 m-1-&gt;next 指向 n 节点， m-&gt;next 指向n+1 节点</span></span><br><span class="line">        m_1-&gt;next = nn;</span><br><span class="line">        mm-&gt;next = n_1;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> dummy-&gt;next;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="234-回文链表"><a href="#234-回文链表" class="headerlink" title="234. 回文链表"></a>234. 回文链表</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/111.png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>快慢双指针结合反转链表：</strong>先走，（快指针走两步，慢指针走一步）若节点个数为<strong>偶数</strong>，快指针最终会指向 NULL 位置；若节点个数为<strong>奇数</strong>，快指针最终会指向最后一个位置而不是 NULL；慢指针则会指向中间位置；</li>
<li>将快指针指向链表头部；将慢指针的链表反转；此时慢指针回指向反转链表的头部</li>
<li>然后移动两个指针，一次移动一步，对比是否相等</li>
</ol>
<ul>
<li>结果</li>
</ul>
<p>时间复杂度 O(n) 	空间复杂度 O(1) 	</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">isPalindrome</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 定义快慢指针</span></span><br><span class="line">        ListNode *fast_node = head;</span><br><span class="line">        ListNode *slow_node = head;</span><br><span class="line">        <span class="comment">// 移动，快指针移动两步，慢指针移动一步</span></span><br><span class="line">        <span class="keyword">while</span>(fast_node-&gt;next != <span class="literal">NULL</span> &amp;&amp; fast_node-&gt;next-&gt;next != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            fast_node = fast_node-&gt;next-&gt;next;</span><br><span class="line">            slow_node = slow_node-&gt;next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 分两种情况，当节点为奇数个时，把中间位置归到左边；偶数个时，刚好是中间的位置</span></span><br><span class="line">        <span class="keyword">if</span>(fast_node != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;   </span><br><span class="line">            slow_node = slow_node-&gt;next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 反转slow_node，slow_node刚好指向尾元素</span></span><br><span class="line">        slow_node = <span class="built_in">reverseList</span>(slow_node);</span><br><span class="line">        <span class="comment">// 快指针指向头</span></span><br><span class="line">        fast_node = head;</span><br><span class="line">        <span class="comment">// 头指针往右，尾指针往左，对比两个链表是否完全相等</span></span><br><span class="line">        <span class="keyword">while</span>(slow_node != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 每次判断两个指针是否相等，再移动节点</span></span><br><span class="line">            <span class="keyword">if</span>(fast_node-&gt;val != slow_node-&gt;val)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 如果两个指针没有相遇，说明不是回文链表</span></span><br><span class="line">                <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">            &#125;</span><br><span class="line">            fast_node = fast_node-&gt;next;</span><br><span class="line">            slow_node = slow_node-&gt;next;</span><br><span class="line"></span><br><span class="line">        &#125;   </span><br><span class="line">        <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="function">ListNode* <span class="title">reverseList</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 上一个节点</span></span><br><span class="line">        ListNode *pre = <span class="literal">NULL</span>;</span><br><span class="line">        <span class="comment">// 当前节点</span></span><br><span class="line">        ListNode *cur = head;</span><br><span class="line">        <span class="comment">// 遍历节点，使得每一次当前节点指向上一个节点</span></span><br><span class="line">        <span class="keyword">while</span>(cur != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 记录下一个节点</span></span><br><span class="line">            ListNode *next = cur-&gt;next;</span><br><span class="line">            <span class="comment">// 指向上一个节点</span></span><br><span class="line">            cur-&gt;next = pre;</span><br><span class="line">            <span class="comment">// 更新上一个节点</span></span><br><span class="line">            pre = cur;</span><br><span class="line">            <span class="comment">// 更新当前节点</span></span><br><span class="line">            cur = next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> pre;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="876-链表的中间节点"><a href="#876-链表的中间节点" class="headerlink" title="876. 链表的中间节点"></a>876. 链表的中间节点</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/$R%5D%60L~YQ8ZXXKXIF5@LQS9O.png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><p><strong>遍历法：</strong>先遍历一遍链表得到节点个数；然后再一次遍历到（n&#x2F;2 + 1）的位置即可；</p>
</li>
<li><p><strong>快慢指针法：</strong>（快指针是慢指针移动速度的两倍）当快指针移动结束时，慢指针移动的位置刚好满足要求</p>
</li>
</ol>
<ul>
<li>结果</li>
</ul>
<p>时间复杂度 O(n) 	空间复杂度 O(1) 	</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">middleNode</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">        ListNode *fast_node = head;</span><br><span class="line">        ListNode *slow_node = head;</span><br><span class="line">        <span class="comment">// 移动的前提是下下个节点存在</span></span><br><span class="line">        <span class="keyword">while</span> (fast_node-&gt;next != <span class="literal">NULL</span> &amp;&amp; fast_node-&gt;next-&gt;next != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 快指针移动两步</span></span><br><span class="line">            fast_node = fast_node-&gt;next-&gt;next;</span><br><span class="line">            <span class="comment">// 满指针移动一步</span></span><br><span class="line">            slow_node = slow_node-&gt;next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 移动结束，直接返回满指针即可</span></span><br><span class="line">        <span class="keyword">return</span> slow_node;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="19-删除链表中的倒数第-n-个节点"><a href="#19-删除链表中的倒数第-n-个节点" class="headerlink" title="19. 删除链表中的倒数第 n 个节点"></a>19. 删除链表中的倒数第 n 个节点<img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/sad.png"></h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>遍历法：</strong>第一次遍历得到链表长度；第二次遍历到第 n-k+1 的位置， 将其删除</li>
<li><strong>哈希表法</strong>：第一次遍历将所有节点存入哈希表（key&#x3D;i，value&#x3D;节点）中，并得到链表长度；然后直接查表 key &#x3D; n-k+1 的值；</li>
<li><strong>固定步长快慢双指针法：</strong>1. 建立虚拟节点； 2. 定义一个快指针，先走 N 步； 3. 快慢指针同时往后走；4. 最终快指针走到最后一个节点，此时的慢指针刚好走到倒数的 n + 1 个节点处；5. 连接n + 1 和 n - 1 两个节点</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">removeNthFromEnd</span><span class="params">(ListNode* head, <span class="type">int</span> n)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 1. 创建虚拟节点</span></span><br><span class="line">        <span class="keyword">auto</span> dummy = <span class="keyword">new</span> <span class="built_in">ListNode</span>(<span class="number">-1</span>);</span><br><span class="line">        <span class="comment">// 虚拟节点的子节点指向头</span></span><br><span class="line">        dummy-&gt;next = head;</span><br><span class="line">        <span class="comment">// 2. 定义快慢指针,指向虚拟节点</span></span><br><span class="line">        <span class="keyword">auto</span> fast_node = dummy, slow_node = dummy;</span><br><span class="line">        <span class="comment">// 3. 快指针先走 n 步</span></span><br><span class="line">        <span class="keyword">while</span>(n--) fast_node = fast_node-&gt;next;</span><br><span class="line">        <span class="comment">// 4. 快慢指针开始同时走,当快指针走到最后一个节点时停止</span></span><br><span class="line">        <span class="keyword">while</span>(fast_node-&gt;next != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            fast_node = fast_node-&gt;next;</span><br><span class="line">            slow_node = slow_node-&gt;next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 此时慢指针就是处于倒数 n+1 的位置</span></span><br><span class="line">        slow_node-&gt;next = slow_node-&gt;next-&gt;next;</span><br><span class="line">        <span class="keyword">return</span> dummy-&gt;next;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="2-两数相加"><a href="#2-两数相加" class="headerlink" title="2. 两数相加"></a>2. 两数相加</h4><ul>
<li>程序</li>
</ul>
<ol>
<li>需要创建一个新的链表用来返回目标值；同时需要创建一个头节点来记录头位置，这样最后才能访问到第一个位置；</li>
<li>每次创建一个临时链表，将计算完的数值存储完之后，再连接到主链表上；</li>
<li>链表结尾如果还需进位，需要最后再加一个节点补上；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">addTwoNumbers</span><span class="params">(ListNode* l1, ListNode* l2)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御编码  </span></span><br><span class="line">        <span class="keyword">if</span>(l1 == <span class="literal">NULL</span> &amp;&amp; l2 == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">NULL</span>;</span><br><span class="line">        ListNode* cur = <span class="keyword">new</span> <span class="built_in">ListNode</span>();</span><br><span class="line">        <span class="comment">// 记住当前链表头的位置</span></span><br><span class="line">        ListNode* head = cur;</span><br><span class="line">        <span class="comment">// 进位标志位</span></span><br><span class="line">        <span class="type">bool</span> flag = <span class="literal">false</span>;</span><br><span class="line">        <span class="type">int</span> temp;</span><br><span class="line">        <span class="comment">// 两个链表不空的情况下进行计算</span></span><br><span class="line">        <span class="keyword">while</span>(l1 != <span class="literal">NULL</span> || l2 != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;   </span><br><span class="line">            <span class="comment">// 判断两个节点是否空，空的话值就取0；不空就取它的值</span></span><br><span class="line">            temp = (l1 != <span class="literal">NULL</span> ? l1-&gt;val : <span class="number">0</span>) + (l2 != <span class="literal">NULL</span> ? l2-&gt;val : <span class="number">0</span>);</span><br><span class="line">            <span class="comment">// 取出值后，转到下一节点</span></span><br><span class="line">            <span class="keyword">if</span>(l1 != <span class="literal">NULL</span>) l1 = l1-&gt;next;</span><br><span class="line">            <span class="keyword">if</span>(l2 != <span class="literal">NULL</span>) l2 = l2-&gt;next;</span><br><span class="line"></span><br><span class="line">            <span class="comment">// 需进位，数值+1，进位标志置为false</span></span><br><span class="line">            <span class="keyword">if</span>(flag)</span><br><span class="line">            &#123;</span><br><span class="line">                temp += <span class="number">1</span>; </span><br><span class="line">                flag = <span class="literal">false</span>;</span><br><span class="line">            &#125; </span><br><span class="line">            <span class="comment">// 再判断是不是大于等于10，进下一位</span></span><br><span class="line">            <span class="keyword">if</span>(temp &gt;=<span class="number">10</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                 temp -=<span class="number">10</span>; </span><br><span class="line">                 flag = <span class="literal">true</span>;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 新建一个节点，连接上目标节点</span></span><br><span class="line">            ListNode* next = <span class="keyword">new</span> <span class="built_in">ListNode</span>();</span><br><span class="line">            next-&gt;val = temp;</span><br><span class="line">            cur-&gt;next = next;</span><br><span class="line">            <span class="comment">// 更新当前节点的位置</span></span><br><span class="line">            cur = next;</span><br><span class="line">            temp = <span class="number">0</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 如果最后一个节点还有进位，那么需要再加一个节点；</span></span><br><span class="line">        <span class="keyword">if</span>(flag) </span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 新建一个节点，连接上</span></span><br><span class="line">            ListNode* next = <span class="keyword">new</span> <span class="built_in">ListNode</span>();</span><br><span class="line">            next-&gt;val = <span class="number">1</span>;</span><br><span class="line">            cur-&gt;next = next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> head-&gt;next;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="237-删除链表中的节点"><a href="#237-删除链表中的节点" class="headerlink" title="237. 删除链表中的节点"></a>237. 删除链表中的节点</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>先复制下一个节点的值为当前节点值</li>
<li>然后将当前节点的子节点指向子子节点，</li>
<li>这样在数组的层面上就把这个值给删除掉了</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">deleteNode</span><span class="params">(ListNode* node)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 赋值下一个节点的值为当前值</span></span><br><span class="line">        node-&gt;val = node-&gt;next-&gt;val;</span><br><span class="line">        <span class="comment">// 当前节点的子节点指向子子节点</span></span><br><span class="line">        node-&gt;next = node-&gt;next-&gt;next;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="61-旋转链表"><a href="#61-旋转链表" class="headerlink" title="61. 旋转链表"></a>61. 旋转链表</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>先走到倒数第 k + 1 的节点；</li>
<li>保存倒数第 k + 1 以后的节点</li>
<li>将头尾连接起来，也就是快指针的子节点和head 连接；</li>
<li>断开倒数第  k + 1的节点</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">rotateRight</span><span class="params">(ListNode* head, <span class="type">int</span> k)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(head == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">NULL</span>;</span><br><span class="line">        <span class="type">int</span> n = <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">auto</span> h = head;</span><br><span class="line">        <span class="keyword">while</span>(h-&gt;next)</span><br><span class="line">        &#123;</span><br><span class="line">            h = h-&gt;next; </span><br><span class="line">            n++;</span><br><span class="line">        &#125; </span><br><span class="line">        k = k % n;</span><br><span class="line">        <span class="comment">// 先走到倒数第 k+1 的节点</span></span><br><span class="line">        <span class="keyword">auto</span> fast = head , slow = head;</span><br><span class="line">        <span class="keyword">while</span>(k--) fast = fast-&gt;next;</span><br><span class="line">        <span class="keyword">while</span>(fast-&gt;next)</span><br><span class="line">        &#123;</span><br><span class="line">            fast = fast-&gt;next;</span><br><span class="line">            slow = slow-&gt;next;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 把快指针的子节点指向头；</span></span><br><span class="line">        fast-&gt;next = head;</span><br><span class="line">        <span class="comment">// 保存慢指针的节点，</span></span><br><span class="line">        head = slow-&gt;next;</span><br><span class="line">        <span class="comment">// 再将倒数k+1的节点的子节点指向空；</span></span><br><span class="line">        slow-&gt;next = <span class="literal">NULL</span>;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">return</span> head;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="24-两两交换链表中的节点"><a href="#24-两两交换链表中的节点" class="headerlink" title="24. 两两交换链表中的节点"></a>24. 两两交换链表中的节点</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.28(11.22).png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>需要创建一个虚拟节点；</li>
<li>按照上图的指针执行顺序 1 2 3 执行；</li>
<li>如果按照 1 3 2 执行，需要保存 b 节点，得而外开辟内存；</li>
<li>（1. 创建虚拟节点，并将虚拟节点的子节点指向头head （2.  初始化 p，a，b 三个指针（3. 按照顺序执行1 2 3</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">ListNode* <span class="title">swapPairs</span><span class="params">(ListNode* head)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 创建虚拟节点</span></span><br><span class="line">        <span class="keyword">auto</span> dummy = <span class="keyword">new</span> <span class="built_in">ListNode</span>(<span class="number">-1</span>);</span><br><span class="line">        dummy-&gt;next = head;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> p = dummy; p-&gt;next &amp;&amp; p-&gt;next-&gt;next;)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 初始化 a b 节点位置</span></span><br><span class="line">            <span class="keyword">auto</span> a = p-&gt;next, b = p-&gt;next-&gt;next;</span><br><span class="line">            <span class="comment">// 1</span></span><br><span class="line">            p-&gt;next = b;</span><br><span class="line">            <span class="comment">// 2</span></span><br><span class="line">            a-&gt;next = b-&gt;next;</span><br><span class="line">            <span class="comment">// 3</span></span><br><span class="line">            b-&gt;next = a;</span><br><span class="line">            p = a;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> dummy-&gt;next;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h3 id="栈与队列（2）"><a href="#栈与队列（2）" class="headerlink" title="栈与队列（2）"></a>栈与队列（2）</h3><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/%E6%A0%88%E5%92%8C%E9%98%9F%E5%88%97.png"></p>
<p>栈（堆栈）：先进后出   放入和取出的部位是：栈顶</p>
<p>队列：先进先出	队尾进，队头出</p>
<h4 id="232-用栈实现队列"><a href="#232-用栈实现队列" class="headerlink" title="232.  用栈实现队列"></a>232.  用栈实现队列</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>创建两个栈（输入栈、输出栈）；从输入栈的插入元素，取出元素时把输入栈的元素放进输出栈，再取出；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">MyQueue</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="comment">// 创建两个栈（输入栈、输出栈）</span></span><br><span class="line">    stack&lt;<span class="type">int</span>&gt; in;</span><br><span class="line">    stack&lt;<span class="type">int</span>&gt; out;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">MyQueue</span>() &#123;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">push</span><span class="params">(<span class="type">int</span> x)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 插入时放入输入栈</span></span><br><span class="line">        in.<span class="built_in">push</span>(x);</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">pop</span><span class="params">()</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 首先判断输出栈是否为空，不空的话就直接取；空的话要把输入栈的数据传过来</span></span><br><span class="line">        <span class="keyword">if</span>(out.<span class="built_in">empty</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="built_in">in2out</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 先拿到该元素的值</span></span><br><span class="line">        <span class="type">int</span> top = out.<span class="built_in">top</span>();</span><br><span class="line">        <span class="comment">// 再删除</span></span><br><span class="line">        out.<span class="built_in">pop</span>();</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> top;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">peek</span><span class="params">()</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 首先判断输出栈是否为空，不空的话就直接取；空的话要把输入栈的数据传过来</span></span><br><span class="line">        <span class="keyword">if</span>(out.<span class="built_in">empty</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="built_in">in2out</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> out.<span class="built_in">top</span>();</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">empty</span><span class="params">()</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(in.<span class="built_in">empty</span>() &amp;&amp; out.<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        <span class="keyword">else</span> <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">in2out</span><span class="params">()</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">while</span>(!in.<span class="built_in">empty</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 查看 in 栈的头元素，并添加到 out 栈</span></span><br><span class="line">            out.<span class="built_in">push</span>(in.<span class="built_in">top</span>());</span><br><span class="line">            <span class="comment">// 删除 in 站的头元素</span></span><br><span class="line">            in.<span class="built_in">pop</span>();</span><br><span class="line">        &#125;   </span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="394-字符串解码（❗❗❗）"><a href="#394-字符串解码（❗❗❗）" class="headerlink" title="394. 字符串解码（❗❗❗）"></a>394. 字符串解码（❗❗❗）</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>由内到外展开：类似栈的结构</li>
<li>按顺序将字符串压入栈中，如果是字母或’[‘则直接放入栈中直到遇到’]’，当遇到’]’时开始出栈直到’[‘</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">string <span class="title">decodeString</span><span class="params">(string s)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 创建一个数字栈</span></span><br><span class="line">        stack&lt;<span class="type">int</span>&gt; nums;</span><br><span class="line">        <span class="comment">// 创建一个字符串栈</span></span><br><span class="line">        stack&lt;string&gt; st;</span><br><span class="line">        <span class="comment">// 临时数字和字符串</span></span><br><span class="line">        <span class="type">int</span> num = <span class="number">0</span>;</span><br><span class="line">        string tmp;</span><br><span class="line">        <span class="comment">// 字符串长度</span></span><br><span class="line">        <span class="type">int</span> n = s.<span class="built_in">size</span>();</span><br><span class="line">        <span class="comment">// 遍历每一个字符</span></span><br><span class="line">        <span class="keyword">for</span> (<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 如果是数字，则先临时记录下来</span></span><br><span class="line">            <span class="keyword">if</span> (s[i] &gt;= <span class="string">&#x27;0&#x27;</span> &amp;&amp; s[i] &lt;= <span class="string">&#x27;9&#x27;</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                num = <span class="number">10</span> * num + s[i] - <span class="string">&#x27;0&#x27;</span>;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 如果是字符，也临时记录下来</span></span><br><span class="line">            <span class="keyword">else</span> <span class="keyword">if</span> ((s[i] &gt;= <span class="string">&#x27;a&#x27;</span> &amp;&amp; s[i] &lt;= <span class="string">&#x27;z&#x27;</span>) || (s[i] &gt;= <span class="string">&#x27;A&#x27;</span> &amp;&amp; s[i] &lt;= <span class="string">&#x27;Z&#x27;</span>))</span><br><span class="line">            &#123;</span><br><span class="line">                tmp.<span class="built_in">push_back</span>(s[i]);</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 如果是左括号 &#x27;[&#x27;</span></span><br><span class="line">            <span class="keyword">else</span> <span class="keyword">if</span> (s[i] == <span class="string">&#x27;[&#x27;</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 把临时数字和临时字符串压入栈中，先把数字压入栈内，再把字母压入栈内</span></span><br><span class="line">                nums.<span class="built_in">push</span>(num);</span><br><span class="line">                st.<span class="built_in">push</span>(tmp);</span><br><span class="line">                <span class="comment">// 清空临时数字和字符串</span></span><br><span class="line">                num = <span class="number">0</span>;</span><br><span class="line">                tmp.<span class="built_in">clear</span>();</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 如果是右括号 &#x27;]&#x27;</span></span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 取出数字栈的栈顶元素</span></span><br><span class="line">                <span class="type">int</span> cnt = nums.<span class="built_in">top</span>();</span><br><span class="line">                nums.<span class="built_in">pop</span>();</span><br><span class="line">                <span class="comment">// 得到字符串栈的栈顶元素，并将字符串累加 cnt 次）</span></span><br><span class="line">                <span class="keyword">for</span> (<span class="type">int</span> j = <span class="number">0</span>; j &lt; cnt; j++)</span><br><span class="line">                &#123;</span><br><span class="line">                    st.<span class="built_in">top</span>() += tmp;</span><br><span class="line">                &#125;</span><br><span class="line">                <span class="comment">// 得到字符串</span></span><br><span class="line">                tmp = st.<span class="built_in">top</span>();</span><br><span class="line">                <span class="comment">// 移出字符串</span></span><br><span class="line">                st.<span class="built_in">pop</span>();</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> tmp;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h3 id="树（7）"><a href="#树（7）" class="headerlink" title="树（7）"></a>树（7）</h3><p> 满二叉树：每个节点都有左右两个子节点，叶子节点都在同一层</p>
<p> 完全二叉树：序号和满二叉树相同</p>
<ul>
<li>位置为  k 的节点，左子节点：2<em>k + 1，右子节点：2</em>k+2</li>
</ul>
<h4 id="94-二叉树的中序遍历"><a href="#94-二叉树的中序遍历" class="headerlink" title="94. 二叉树的中序遍历"></a>94. 二叉树的中序遍历</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.07.10(4.20).png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<p><strong>顺序：</strong>左节 -&gt; 根节点 -&gt; 右节点</p>
<p><strong>节点访问顺序：</strong>G D H B A E I C F</p>
<ol>
<li><strong>递归：</strong>首先判断当前节点是否为空；</li>
<li>然后判断是否有左节点；（一直到最后一个左子节点，再开始添加元素）</li>
<li>再判断是否有右节点；</li>
<li><strong>循环迭代：</strong>一直压栈直到找到最后一个左子树才开始存值，并出栈依次</li>
<li>用一个栈来存储经过的节点；</li>
<li>首先遍历节点，将左节点压入栈中；</li>
<li>如果当前节点为空，就从栈中拿出刚存入的节点（上一个节点），也就是当前节点的根节点，存储该节点的值；再去访问其右节点；再将该节点存入栈中 ；循环第7步…</li>
</ol>
<ul>
<li>递归</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">inorderTraversal</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; v;</span><br><span class="line">        <span class="built_in">accessTree</span>(root, v);</span><br><span class="line">        <span class="keyword">return</span> v;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">accessTree</span><span class="params">(TreeNode* root, vector&lt;<span class="type">int</span>&gt;&amp; v)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 如果该节点为空</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span>;</span><br><span class="line">        <span class="comment">// 否则，判断是否有左节点</span></span><br><span class="line">        <span class="built_in">accessTree</span>(root-&gt;left, v);</span><br><span class="line">        <span class="comment">// 如果没有左节点，就把根节点添加（就是自己）</span></span><br><span class="line">        v.<span class="built_in">push_back</span>(root-&gt;val);</span><br><span class="line">        <span class="comment">// 再去访问右节点</span></span><br><span class="line">        <span class="built_in">accessTree</span>(root-&gt;right, v);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>迭代</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">inorderTraversal</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; v;</span><br><span class="line">        <span class="comment">// 创建一个栈，用来临时存放二叉树节点,这样就可以访问上一个节点的情况；</span></span><br><span class="line">        stack&lt;TreeNode*&gt; stack;</span><br><span class="line">        <span class="comment">// 1. 创建一个循环，用来遍历所有元素</span></span><br><span class="line">        <span class="keyword">while</span>(root != <span class="literal">NULL</span> || !stack.<span class="built_in">empty</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 2. 创建一个循环，用来遍历左子树</span></span><br><span class="line">            <span class="keyword">while</span>(root != <span class="literal">NULL</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 把左子树压入栈中</span></span><br><span class="line">                stack.<span class="built_in">push</span>(root);</span><br><span class="line">                <span class="comment">// 继续下一个左子树</span></span><br><span class="line">                root = root-&gt;left;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 如果没有左子树，将该节点视为根节点，将该节点的值添加进数组中</span></span><br><span class="line">            root = stack.<span class="built_in">top</span>();</span><br><span class="line">            stack.<span class="built_in">pop</span>();</span><br><span class="line">            v.<span class="built_in">push_back</span>(root-&gt;val);</span><br><span class="line">            <span class="comment">// 再去访问该节点的右子树</span></span><br><span class="line">            root = root-&gt;right;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> v;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="144-二叉树的前序遍历"><a href="#144-二叉树的前序遍历" class="headerlink" title="144. 二叉树的前序遍历"></a>144. 二叉树的前序遍历</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.07.10(4.20).png"></p>
<p><strong>顺序：</strong>根节点 -&gt; 左节点 -&gt; 右节点</p>
<p><strong>节点访问顺序：</strong>A B D G H C E I F</p>
<p>存值：找到根节点就存</p>
<ul>
<li>递归</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">preorderTraversal</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; v;</span><br><span class="line">        <span class="built_in">accessTree</span>(root, v);</span><br><span class="line">        <span class="keyword">return</span> v;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="comment">// 递归</span></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">accessTree</span><span class="params">(TreeNode* root, vector&lt;<span class="type">int</span>&gt;&amp; v)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 访问根节点，空的话就返回</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span>;</span><br><span class="line">        <span class="comment">// 存入根节点</span></span><br><span class="line">        v.<span class="built_in">push_back</span>(root-&gt;val);</span><br><span class="line">        <span class="comment">// 访问左子树</span></span><br><span class="line">        <span class="built_in">accessTree</span>(root-&gt;left, v);</span><br><span class="line">        <span class="comment">// 访问右子树</span></span><br><span class="line">        <span class="built_in">accessTree</span>(root-&gt;right, v);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>迭代</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">preorderTraversal</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; v;</span><br><span class="line">        <span class="comment">// 创建一个栈 用来存root</span></span><br><span class="line">        stack&lt;TreeNode*&gt; stack;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">while</span>(!stack.<span class="built_in">empty</span>() || root != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">while</span>(root != <span class="literal">NULL</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 记录根节点值</span></span><br><span class="line">                v.<span class="built_in">push_back</span>(root-&gt;val);</span><br><span class="line">                <span class="comment">// 将该根节点存入栈中</span></span><br><span class="line">                stack.<span class="built_in">push</span>(root);</span><br><span class="line">                <span class="comment">// 继续访问左节点</span></span><br><span class="line">                root = root-&gt;left;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 取出栈中的根节点</span></span><br><span class="line">            root = stack.<span class="built_in">top</span>();</span><br><span class="line">            stack.<span class="built_in">pop</span>();</span><br><span class="line">            <span class="comment">// 找根节点的右节点</span></span><br><span class="line">            root = root-&gt;right;</span><br><span class="line">        &#125;   </span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> v;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h4 id="145-二叉树的后序遍历"><a href="#145-二叉树的后序遍历" class="headerlink" title="145. 二叉树的后序遍历"></a>145. 二叉树的后序遍历</h4><p><strong>顺序：</strong>左节点 -&gt; 右节点 -&gt; 根节点</p>
<p><strong>节点访问顺序：</strong>G H D B I E F C A</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>创建一个变量：用来判断之前是否走过右节点</li>
<li>左节点保存完之后，从栈中取出根节点，如果其右节点存在，则再把根节点压回栈中，以便下一次访问；然后再去访问存在的右节点，将右节点存入栈中；当判断完右节点后，并将右节点的值保存。再取出根节点，通过判断标志位，来知道之前是否走过，走过的话直接跳到下一个节点；否则继续访问右节点；</li>
</ol>
<p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.07.10(4.20).png"></p>
<ul>
<li>递归</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">postorderTraversal</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; v;</span><br><span class="line">        <span class="built_in">accessTree</span>(root, v);</span><br><span class="line">        <span class="keyword">return</span> v;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="comment">// 递归</span></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">accessTree</span><span class="params">(TreeNode* root, vector&lt;<span class="type">int</span>&gt;&amp; v)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 首先判断该节点是否为空，空的话返回上一层递归</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span>;</span><br><span class="line">        <span class="comment">// 如果非空，则访问左节点</span></span><br><span class="line">        <span class="built_in">accessTree</span>(root-&gt;left, v);</span><br><span class="line">        <span class="comment">// 再访问右节点</span></span><br><span class="line">        <span class="built_in">accessTree</span>(root-&gt;right, v);</span><br><span class="line">        <span class="comment">// 存入该值</span></span><br><span class="line">        v.<span class="built_in">push_back</span>(root-&gt;val);</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>迭代</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">postorderTraversal</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        vector&lt;<span class="type">int</span>&gt; v;</span><br><span class="line">        stack&lt;TreeNode*&gt; stack;</span><br><span class="line">        <span class="comment">// 用来判断是否有走过；</span></span><br><span class="line">        TreeNode* pre = <span class="literal">NULL</span>;</span><br><span class="line">        <span class="keyword">while</span>(!stack.<span class="built_in">empty</span>() || root != <span class="literal">NULL</span>)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">while</span>(root != <span class="literal">NULL</span>)</span><br><span class="line">            &#123;   </span><br><span class="line">                <span class="comment">// 将节点存入栈中</span></span><br><span class="line">                stack.<span class="built_in">push</span>(root);</span><br><span class="line">                <span class="comment">// 继续访问左节点</span></span><br><span class="line">                root = root-&gt;left;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 如果左节点为空，则得取出其根节点，再去访问右节点；</span></span><br><span class="line">            root = stack.<span class="built_in">top</span>();</span><br><span class="line">            stack.<span class="built_in">pop</span>();</span><br><span class="line">            <span class="comment">// 若右节点为空，则将根节点值存入</span></span><br><span class="line">            <span class="keyword">if</span>(root-&gt;right == <span class="literal">NULL</span> || root-&gt;right == pre)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 存值</span></span><br><span class="line">                v.<span class="built_in">push_back</span>(root-&gt;val);</span><br><span class="line">                pre = root;</span><br><span class="line">                <span class="comment">// root 置为空，才能再取出上一节点</span></span><br><span class="line">                root = <span class="literal">NULL</span>;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 若右节点非空</span></span><br><span class="line">            <span class="keyword">else</span></span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 重新将主节点压入栈中，确保下一次从栈从还能访问到其根节点</span></span><br><span class="line">                stack.<span class="built_in">push</span>(root);</span><br><span class="line">                root = root-&gt;right;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> v;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="101-对称二叉树"><a href="#101-对称二叉树" class="headerlink" title="101. 对称二叉树"></a>101. 对称二叉树</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>对称的条件：</strong>左节点的左子树 &#x3D;&#x3D; 右节点的右子树</li>
<li>左节点的右子树 &#x3D;&#x3D; 右节点的左子树</li>
<li><strong>递归法：</strong>递归的终止条件有三种情况：3.1 左右两边的节点都相等为空，则返回true对称；3.2 左右两边节点有一个为空，则返回 false不对称；3.3 左右两边都不空，但是值不相等，则返回false不对称；</li>
<li>否则继续递归同时判断：左节点的左子树-右节点的右子树；左节点的右子树-右节点的左子树；</li>
</ol>
<p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.12.png"></p>
<ol>
<li><strong>循环迭代：</strong>借助队列实现；</li>
<li>初始条件：如果第一个root节点为空，或其左右节点为空，则必是对称；否则将左右子节点存入队列中；然后从队列中取出两个节点，判断左节点是否等于右节点（包括空和非空但是值不相等和相等三种情况）；</li>
<li>是的话，依次讲左的左子树-&gt;右的右子树-&gt;左的右子树-&gt; 右的左子树存入队列中；继续下一次再取出两个节点判断；</li>
<li>不是的话，就不是对称的；</li>
</ol>
<p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.12%EF%BC%885.29%EF%BC%89.png"></p>
<ul>
<li>递归方法</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">isSymmetric</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">accessTree</span>(root-&gt;left, root-&gt;right);</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line"></span><br><span class="line">    <span class="comment">// 递归(需判断左右两个节点)</span></span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">accessTree</span><span class="params">(TreeNode* left, TreeNode* right)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 如果左右节点都为空，则是对称的</span></span><br><span class="line">        <span class="keyword">if</span>(left == <span class="literal">NULL</span> &amp;&amp; right == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        <span class="comment">// 如果左右节点其中一个空，则不是对称的</span></span><br><span class="line">        <span class="keyword">if</span>(left == <span class="literal">NULL</span> || right == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">        <span class="comment">// 如果左右节点都不空，但是值不等，则不是对称的</span></span><br><span class="line">        <span class="keyword">if</span>(left-&gt;val != right-&gt;val) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">        <span class="comment">// 否则下一次判断</span></span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">accessTree</span>(left-&gt;left, right-&gt;right) &amp;&amp; <span class="built_in">accessTree</span>(left-&gt;right, right-&gt;left);</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<ul>
<li>循环迭代</li>
</ul>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">isSymmetric</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        TreeNode* l;</span><br><span class="line">        TreeNode* r;</span><br><span class="line">        queue&lt;TreeNode*&gt; queue;</span><br><span class="line">        <span class="comment">// 初始条件</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span> || (root-&gt;left==<span class="literal">NULL</span>&amp;&amp; root-&gt;right==<span class="literal">NULL</span>)) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        <span class="comment">// 将左右两个节点存入队列中</span></span><br><span class="line">        queue.<span class="built_in">push</span>(root-&gt;left);</span><br><span class="line">        queue.<span class="built_in">push</span>(root-&gt;right);</span><br><span class="line">        <span class="keyword">while</span>(!queue.<span class="built_in">empty</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 从队列中取出两个元素</span></span><br><span class="line">            l = queue.<span class="built_in">front</span>();</span><br><span class="line">            queue.<span class="built_in">pop</span>();</span><br><span class="line">            r = queue.<span class="built_in">front</span>();</span><br><span class="line">            queue.<span class="built_in">pop</span>();</span><br><span class="line">            <span class="comment">// 判断是否对称条件：</span></span><br><span class="line">            <span class="comment">// 1. 判断是否为空，为空的话继续判断下一个;</span></span><br><span class="line">            <span class="keyword">if</span>(l == <span class="literal">NULL</span> &amp;&amp; r == <span class="literal">NULL</span>) <span class="keyword">continue</span>;</span><br><span class="line">            <span class="comment">// 2. 判断是否相等，不相等的话直接返回false;</span></span><br><span class="line">            <span class="keyword">if</span>(l == <span class="literal">NULL</span> || r == <span class="literal">NULL</span> || l-&gt;val != r-&gt;val) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">            </span><br><span class="line">            <span class="comment">// 将这四个节点存入队列中</span></span><br><span class="line">            queue.<span class="built_in">push</span>(l-&gt;left);</span><br><span class="line">            queue.<span class="built_in">push</span>(r-&gt;right);</span><br><span class="line">            queue.<span class="built_in">push</span>(l-&gt;right);</span><br><span class="line">            queue.<span class="built_in">push</span>(r-&gt;left);</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="104-二叉树的最大深度"><a href="#104-二叉树的最大深度" class="headerlink" title="104. 二叉树的最大深度"></a>104. 二叉树的最大深度</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>递归法：</strong>终止条件：该节点为NULL时，返回0;</li>
<li>否则返回，其左右子节点的深度更大的那个的值+1；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="comment">// 递归</span></span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">maxDepth</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 该节点的深度为0;</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">max</span>(<span class="built_in">maxDepth</span>(root-&gt;left), <span class="built_in">maxDepth</span>(root-&gt;right)) + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="110-平衡二叉树"><a href="#110-平衡二叉树" class="headerlink" title="110. 平衡二叉树"></a>110. 平衡二叉树</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>一颗高度平衡的二叉树定义为：<strong>每个</strong>左右子树的高度差的绝对值不超过1；</li>
<li>每到一个节点都要判断这颗二叉树是否是平衡的；</li>
<li>用迭代法来计算每个节点（二叉树）的深度，同时判断其左右子树是否是平衡的，如果不是，就返回 -1；如果是就返回其深度值；并返回上一节点；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">isBalanced</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御编码 </span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        <span class="comment">// 分别计算子树的值</span></span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">accessTree</span>(root) != <span class="number">-1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="comment">// 递归计算子节点的深度</span></span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">accessTree</span><span class="params">(TreeNode* root)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 终止条件</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        <span class="comment">// 计算左子树的深度</span></span><br><span class="line">        <span class="type">int</span> left = <span class="built_in">accessTree</span>(root-&gt;left);</span><br><span class="line">        <span class="type">int</span> right = <span class="built_in">accessTree</span>(root-&gt;right);</span><br><span class="line">        <span class="comment">// 如果有任何一颗二叉树是不平衡的就返回 -1</span></span><br><span class="line">        <span class="keyword">if</span>(<span class="built_in">abs</span>(left - right) &gt; <span class="number">1</span> || left == <span class="number">-1</span> || right == <span class="number">-1</span>) <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">max</span>(left, right) + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="226-翻转二叉树"><a href="#226-翻转二叉树" class="headerlink" title="226. 翻转二叉树"></a>226. 翻转二叉树</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>递归：</strong>一直递归到叶节点，然后再开始一步一步自底向上交换；</li>
<li>先从每个叶节点开始左右交换，然后往上一层交换，直至剩一个；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">TreeNode* <span class="title">invertTree</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御编码</span></span><br><span class="line">        <span class="keyword">if</span>(root == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">NULL</span>;</span><br><span class="line">        <span class="comment">// 先交换其子叶的左右节点</span></span><br><span class="line">        <span class="built_in">invertTree</span>(root-&gt;left);</span><br><span class="line">        <span class="comment">// 再交换子叶节点</span></span><br><span class="line">        <span class="built_in">invertTree</span>(root-&gt;right);</span><br><span class="line">        <span class="comment">// 再交换左右节点</span></span><br><span class="line">        TreeNode* temp = root-&gt;left;</span><br><span class="line">        root-&gt;left = root-&gt;right;</span><br><span class="line">        root-&gt;right = temp;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> root;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="98-验证二叉搜索树"><a href="#98-验证二叉搜索树" class="headerlink" title="98. 验证二叉搜索树"></a>98. 验证二叉搜索树</h4><ul>
<li><strong>思路</strong></li>
</ul>
<p>所有左右子树满足：左节点 &lt; 父节点 &lt; 右节点</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>一个节点一个节点判断，每个节点都有一个取值范围；不要一颗树判断</li>
<li>递归输入为当前节点和节点值的范围</li>
<li>递归的终止条件：1. 如果当前节点为空，说明之前的节点都符合条件，则返回 true； 2. 如果当前节点的值超出了范围，则不满足条件，返回false；</li>
<li>递归的执行函数：dfs(左) &amp;&amp; dfs(右)</li>
<li>左节点更新右区间，右子树跟新左区间；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">isValidBST</span><span class="params">(TreeNode* root)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">dfs</span>(root, INT_MIN, INT_MAX);</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">dfs</span><span class="params">(TreeNode* root, <span class="type">long</span> <span class="type">long</span> minV, <span class="type">long</span> <span class="type">long</span> maxV)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(!root) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        <span class="keyword">if</span>(root-&gt;val &lt; minV || root-&gt;val &gt; maxV) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> <span class="built_in">dfs</span>(root-&gt;left, minV, root-&gt;val - <span class="number">1ll</span>) &amp;&amp; <span class="built_in">dfs</span>(root-&gt;right, root-&gt;val + <span class="number">1ll</span>, maxV);</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h3 id="二分查找（10）"><a href="#二分查找（10）" class="headerlink" title="二分查找（10）"></a>二分查找（10）</h3><p><strong>性质汇总：</strong></p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 用来查目标值（所有序列）以及位置</span></span><br><span class="line">nums[mid] &gt;= target;</span><br><span class="line"><span class="comment">// 用来查最小值（旋转序列）以及位置</span></span><br><span class="line">nums[mid] &lt;= nums.<span class="built_in">back</span>();</span><br></pre></td></tr></table></figure>

<p><strong>性质一定是要符合整个区间的二段性的，不能只符合其中一小段！！！</strong></p>
<p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.15(15.45).png"></p>
<p><strong>模板一：当总条件是右边时</strong></p>
<p>当我们将区间 [l, r]  划分成 [l, mid] 和 [mid +1, r] 时，更新操作是：r &#x3D; mid 或 l &#x3D; mid + 1; 计算 mid 是不需要 +1；</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="type">int</span> <span class="title">search</span><span class="params">(<span class="type">int</span> left, <span class="type">int</span> right)</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">    <span class="keyword">while</span>(left &lt; right)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="type">int</span> mid = left + right &gt;&gt; <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span>(<span class="built_in">check</span>(mid)) right = mid;</span><br><span class="line">        <span class="keyword">else</span> left = mid + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> left;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p><strong>模板二：当总条件是左边时</strong></p>
<p>当我们将区间 [l, r] 划分成 [l, mid - 1] 和 [mid, r] 时，更新操作是 r &#x3D; mid - 1 或 l &#x3D; mid;此时为了防止死循环，计算 mid 时需要 +1;</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br></pre></td><td class="code"><pre><span class="line"><span class="function"><span class="type">int</span> <span class="title">search</span><span class="params">(<span class="type">int</span> left, <span class="type">int</span> right)</span></span></span><br><span class="line"><span class="function"></span>&#123;</span><br><span class="line">    <span class="keyword">while</span>(left &lt; right)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="type">int</span> mid = left + right + <span class="number">1</span> &gt;&gt; <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">if</span>(<span class="built_in">check</span>(mid)) left = mid;</span><br><span class="line">        <span class="keyword">else</span> right = mid - <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="keyword">return</span> left;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p> <strong>二分法的流程</strong></p>
<ol>
<li>确定二分的边界</li>
<li>编写二分的代码框架</li>
<li>设计一个check函数（性质）</li>
<li>判断一下区间如何更新</li>
<li>如果更新方式写的是：l &#x3D; mid ，r &#x3D; mid - 1 ，那么就在算mid的时候加上1</li>
</ol>
<h4 id="69-X-的平方根"><a href="#69-X-的平方根" class="headerlink" title="69. X 的平方根"></a>69. X 的平方根</h4><ul>
<li>程序</li>
</ul>
<ol>
<li>确定二分边界 （0， x ）</li>
<li>写二分的框架</li>
<li>写性质</li>
<li>判断更新方式，决定mid的更新方式</li>
</ol>
<p>采用性质：<br>$$<br>左边：[ 0, \sqrt x ] ，右边：[\sqrt x+1, x]<br>$$</p>
<p>$$<br>\sqrt x 位于左边区间，因此使用模板二<br>$$</p>
<p>代码：</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">mySqrt</span><span class="params">(<span class="type">int</span> x)</span> </span>&#123;</span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>;</span><br><span class="line">        <span class="type">int</span> r = x;</span><br><span class="line">        <span class="keyword">while</span>(l &lt;  r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// r+1 有可能会溢出</span></span><br><span class="line">            <span class="type">int</span> mid = (<span class="type">long</span> <span class="type">long</span>)l + r + <span class="number">1</span>&gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="comment">// mid*mid 有可能会溢出 </span></span><br><span class="line">            <span class="keyword">if</span>((<span class="type">long</span> <span class="type">long</span> )mid * mid &lt;= x ) l = mid;</span><br><span class="line">            <span class="keyword">else</span> r = mid - <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<p>采用性质：m*m &gt; x<br>$$<br>左边：[ 0, \sqrt x ] ，右边：[\sqrt x+1, x]<br>$$</p>
<p>$$<br>\sqrt x位于左边区间，因此使用模板二<br>$$</p>
<p>代码</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"></span><br><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">mySqrt</span><span class="params">(<span class="type">int</span> x)</span> </span>&#123;</span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>;</span><br><span class="line">        <span class="type">int</span> r = x;</span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = (<span class="type">long</span> <span class="type">long</span>)l + r + <span class="number">1</span>&gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>((<span class="type">long</span> <span class="type">long</span> )mid * mid &gt; x) r = mid - <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">else</span> l = mid ;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="35-搜索插入位置"><a href="#35-搜索插入位置" class="headerlink" title="35. 搜索插入位置"></a>35. 搜索插入位置</h4><ul>
<li>程序</li>
</ul>
<ol>
<li>确定二分边界 （0， nums.size() - 1 ）</li>
<li>写二分的框架</li>
<li>写性质；判断 x 属于左、右哪个区间；属于左区间就选用模板二，右区间就选用模板一</li>
<li>判断更新方式，决定mid的更新方式</li>
</ol>
<p>性质选用 nums[i] &gt;&#x3D; x，x 位于右区间，选用模板一</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">searchInsert</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums, <span class="type">int</span> target)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(nums.<span class="built_in">empty</span>() || nums.<span class="built_in">back</span>() &lt; target) <span class="keyword">return</span> nums.<span class="built_in">size</span>();</span><br><span class="line"></span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = nums.<span class="built_in">size</span>()<span class="number">-1</span>;</span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(nums[mid] &gt;= target) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br></pre></td></tr></table></figure>

<p>性质选用 nums[i] &lt; x，x 位于右区间，选用模板一</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">searchInsert</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums, <span class="type">int</span> target)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(nums.<span class="built_in">empty</span>() || nums.<span class="built_in">back</span>() &lt; target) <span class="keyword">return</span> nums.<span class="built_in">size</span>();</span><br><span class="line"></span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = nums.<span class="built_in">size</span>()<span class="number">-1</span>;</span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(nums[mid] &lt; target) l = mid + <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">else</span> r = mid;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="35-在排序数组中查找元素的第一个和最后一个位置"><a href="#35-在排序数组中查找元素的第一个和最后一个位置" class="headerlink" title="35. 在排序数组中查找元素的第一个和最后一个位置"></a>35. 在排序数组中查找元素的第一个和最后一个位置</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>确定边界 [0, nums.size() - 1]</li>
<li>写二分框架</li>
<li>确定性质为：nums[mid] &gt;&#x3D; target，那么target只出现在右区间，使用模板一；可找出target的第一个位置；如果没找到，说明没有这个数；</li>
<li>当前只找到第一个的位置，还需要找第二个位置；</li>
<li>因此使用性质为：nums[mid] &lt;&#x3D; target，那么target只出现在左区间，使用模板二；可找出target的第二个位置；返回第一个和第二个位置；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">searchRange</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums, <span class="type">int</span> target)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御编码</span></span><br><span class="line">        <span class="keyword">if</span>(nums.<span class="built_in">empty</span>()) <span class="keyword">return</span> &#123;<span class="number">-1</span>, <span class="number">-1</span>&#125;;</span><br><span class="line">        <span class="comment">// 找第一个位置</span></span><br><span class="line">        <span class="comment">// 1. 确定边界</span></span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = nums.<span class="built_in">size</span>() - <span class="number">1</span>;</span><br><span class="line">        <span class="comment">// 2. 二分框架</span></span><br><span class="line">        <span class="keyword">while</span>(l&lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 4. 确定 mid 写法</span></span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="comment">// 3. 找性质</span></span><br><span class="line">            <span class="keyword">if</span>(nums[mid] &gt;= target) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>; </span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 如果二分的结果不等于目标，说明没找到;</span></span><br><span class="line">        <span class="keyword">if</span>(nums[r] != target) <span class="keyword">return</span> &#123;<span class="number">-1</span>, <span class="number">-1</span>&#125;;</span><br><span class="line">        <span class="type">int</span> start = r;</span><br><span class="line">        <span class="comment">// 找第二个位置</span></span><br><span class="line">        l = <span class="number">0</span>, r = nums.<span class="built_in">size</span>() <span class="number">-1</span>;</span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = l + r + <span class="number">1</span> &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(nums[mid] &lt;= target) l = mid;</span><br><span class="line">            <span class="keyword">else</span> r = mid - <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 再次找到第二个</span></span><br><span class="line">        <span class="type">int</span> end = r;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> &#123;start, end&#125;;</span><br><span class="line">    </span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="74-搜索二维矩阵"><a href="#74-搜索二维矩阵" class="headerlink" title="74. 搜索二维矩阵"></a>74. 搜索二维矩阵</h4><ul>
<li>程序</li>
</ul>
<ol>
<li>将二维矩阵转换为一位数组的形式；</li>
<li>矩阵坐标转数组坐标的公式：</li>
</ol>
<p>$$<br>m:矩阵行数 \ n :矩阵列数 \k:坐标，范围间在 [0, m*n -1]<br>$$</p>
<p>k &#x3D; matrix(x &#x2F;  n)(x % n)     除以的是列数！！！</p>
<ol start="3">
<li>二分查找五个步骤</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">searchMatrix</span><span class="params">(vector&lt;vector&lt;<span class="type">int</span>&gt;&gt;&amp; matrix, <span class="type">int</span> target)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 防御编码</span></span><br><span class="line">        <span class="keyword">if</span>(matrix.<span class="built_in">empty</span>() || matrix[<span class="number">0</span>].<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">        <span class="comment">// 矩阵行数</span></span><br><span class="line">        <span class="type">int</span> m = matrix.<span class="built_in">size</span>();</span><br><span class="line">        <span class="comment">// 矩阵列数</span></span><br><span class="line">        <span class="type">int</span> n = matrix[<span class="number">0</span>].<span class="built_in">size</span>(); </span><br><span class="line">        <span class="comment">// 确定区间</span></span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = m * n <span class="number">-1</span>;</span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(matrix[mid / n][mid % n] &gt;= target) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span>(matrix[l / n][l % n] == target) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">        <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="153-寻找旋转排序数组中的最小值"><a href="#153-寻找旋转排序数组中的最小值" class="headerlink" title="153. 寻找旋转排序数组中的最小值"></a>153. 寻找旋转排序数组中的最小值</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.15(17.1).png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>二分找性质：</li>
<li>性质必须包含除了目标之外的其他数，如果符合性质的只有目标值，则这条性质不行</li>
<li>性质：nums[mid] &lt;&#x3D; nums.back()</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">findMin</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = nums.<span class="built_in">size</span>() - <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(nums[mid] &lt;= nums.<span class="built_in">back</span>()) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>; </span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> nums[l];</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="33-搜索旋转排序数组"><a href="#33-搜索旋转排序数组" class="headerlink" title="33. 搜索旋转排序数组"></a>33. 搜索旋转排序数组</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>先用一次二分法（性质：nums[mid] &lt;&#x3D; nums.back()）找到最小值；</li>
<li>判断最小值属于左边区间、还是右边区间</li>
<li>再用一次二分法（性质：nums[mid] &gt;&#x3D; target），找出；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">search</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums, <span class="type">int</span> target)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 1. 第一次二分，确定目标值位于哪个区间</span></span><br><span class="line">        <span class="comment">// 确定左右区间</span></span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = nums.<span class="built_in">size</span>()<span class="number">-1</span>;</span><br><span class="line">        <span class="comment">// 二分框架</span></span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 确定mid</span></span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="comment">// 确定性质:采用模板一，在右区间</span></span><br><span class="line">            <span class="keyword">if</span>(nums[mid] &lt;= nums.<span class="built_in">back</span>()) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 判断目标值属于哪个区间,再更新左右边界</span></span><br><span class="line">        <span class="keyword">if</span>(target &lt;= nums.<span class="built_in">back</span>()) r = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="keyword">else</span> l = <span class="number">0</span>, r--;</span><br><span class="line">        <span class="comment">// 2. 再次二分，用来查这个数</span></span><br><span class="line">        <span class="keyword">while</span>(l &lt; r) </span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt;<span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(nums[mid]&gt;= target) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">if</span>(nums[l] == target) <span class="keyword">return</span> l;</span><br><span class="line">        <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="278-第一个错误版本"><a href="#278-第一个错误版本" class="headerlink" title="278. 第一个错误版本"></a>278. 第一个错误版本</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>确定区间 [0, n]</li>
<li>写框架</li>
<li>确定性质：isBadVersion(mid) &#x3D;&#x3D; True；目标在右区间，模板一</li>
<li>写 mid 形式</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">firstBadVersion</span><span class="params">(<span class="type">int</span> n)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 1. 确定区间 [0, n]</span></span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = n;</span><br><span class="line">        <span class="comment">// 2. 写框架</span></span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 4. 写 mid 形式</span></span><br><span class="line">            <span class="type">int</span> mid = (<span class="type">long</span> <span class="type">long</span>)l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="comment">// 3. 确定性质：isBadVersion(mid) == True；在右区间，模板一</span></span><br><span class="line">            <span class="keyword">if</span>(<span class="built_in">isBadVersion</span>(mid) == <span class="literal">true</span>) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="162-寻找峰值"><a href="#162-寻找峰值" class="headerlink" title="162. 寻找峰值"></a>162. 寻找峰值</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.16(5.26).png"></p>
<ul>
<li><strong>思路</strong></li>
</ul>
<p>选取右区间，如果有 nums[mid] &lt; nums[mid + 1]，说明直线有一段往右上斜，必然存在一个峰值；</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><p>有 nums[mid] &lt; nums[mid + 1]，说明直线有一段往右上斜，必然存在一个峰值，因此取右半段区间；利用模板一；</p>
</li>
<li><p>当存在 nums[mid] &gt; nums[mid + 1]时，说明峰值出现，更新右边界为 mid；</p>
</li>
<li><p>当mid 取到 nums.size()-1 时，此时l &#x3D; r 会直接退出循环，并不会执行 if 语句</p>
</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">findPeakElement</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = nums.<span class="built_in">size</span>() - <span class="number">1</span>;</span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> mid = l + r &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(nums[mid] &gt; nums[mid + <span class="number">1</span>]) r = mid;</span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="287-寻找重复数"><a href="#287-寻找重复数" class="headerlink" title="287.寻找重复数"></a>287.寻找重复数</h4><ul>
<li><strong>思路</strong></li>
</ul>
<p>抽屉原理：n+1个苹果放进 n 个抽屉里，至少会有一个苹果多出来</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>取 [1, n]的中间值 n&#x2F;2 ，判断一次数组中所有 &lt;&#x3D; n&#x2F;2的数有多少个，正常来说应该是：M - L + 1（个）；所有 &gt; n&#x2F;2的数有多少个，正常来说应该是：R - M（个）</li>
<li>所以取一次值，就判断一次个数是否满足条件，不满足就更新那边的区间</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">findDuplicate</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line">        <span class="type">int</span> l = <span class="number">1</span>, r = nums.<span class="built_in">size</span>() <span class="number">-1</span>;</span><br><span class="line">        <span class="keyword">while</span>(l &lt;  r)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// q</span></span><br><span class="line">            <span class="type">int</span> mid = l + (r-l)/<span class="number">2</span>;</span><br><span class="line">            <span class="comment">// 计数</span></span><br><span class="line">            <span class="type">int</span> cnt = <span class="number">0</span>;</span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">auto</span> n: nums)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">if</span>(n &lt;= mid ) cnt++;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 如果在左边</span></span><br><span class="line">            <span class="keyword">if</span>(cnt &gt; mid) r = mid;</span><br><span class="line">            <span class="comment">// 如果在右边</span></span><br><span class="line">            <span class="keyword">else</span> l = mid + <span class="number">1</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="275-H-指数Ⅱ"><a href="#275-H-指数Ⅱ" class="headerlink" title="275. H 指数Ⅱ"></a>275. H 指数Ⅱ</h4><ul>
<li>思路</li>
</ul>
<p>分析：h 的范围是：[0, n]，且具有二分性质</p>
<p>如果 h 满足条件，那么 h - 1 必然也满足；因为数组为升序，既然 &gt;&#x3D;h 的数有 h 个数，那么 &gt;&#x3D; h - 1 的数至少也有 h 个；</p>
<p>因此以 h 为分界点的前半段满足性质；区间位于左边；采用模板二</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>采用模板二</li>
<li>性质：nums[nums.size() - mid] &gt;&#x3D; mid;</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">hIndex</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; citations)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>, r = citations.<span class="built_in">size</span>();</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">while</span>(l &lt; r)</span><br><span class="line">        &#123;</span><br><span class="line"></span><br><span class="line">            <span class="type">int</span> mid = l + r + <span class="number">1</span> &gt;&gt; <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(citations[citations.<span class="built_in">size</span>() - mid] &gt;= mid) l = mid;</span><br><span class="line">            <span class="keyword">else</span> r = mid - <span class="number">1</span>; </span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h3 id="字符串（9）"><a href="#字符串（9）" class="headerlink" title="字符串（9）"></a>字符串（9）</h3><p>常规操作：</p>
<p>查看从某个位置开始，有多少个相同的数；</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; s.<span class="built_in">size</span>(); i++)</span><br><span class="line">&#123;</span><br><span class="line">   	<span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">0</span>; j &lt; s.<span class="built_in">size</span>(); j++)</span><br><span class="line">    &#123;</span><br><span class="line">        <span class="comment">// 从第 j 个位置开始计算有多少个相同的数</span></span><br><span class="line">        <span class="type">int</span> k = j;</span><br><span class="line">        <span class="comment">// 首先保证s[k]不会溢出，再保证遍历到的值相等，这时计数+1；</span></span><br><span class="line">        <span class="keyword">while</span>(k &lt; s.<span class="built_in">size</span>() &amp;&amp; s[j] == s[k]) k++;</span><br><span class="line">        <span class="comment">// 最后计算有几个重复的数</span></span><br><span class="line">        i = k - j;</span><br><span class="line">    &#125; </span><br><span class="line">&#125;</span><br><span class="line"></span><br></pre></td></tr></table></figure>

<h4 id="38-外观数列"><a href="#38-外观数列" class="headerlink" title="38. 外观数列"></a>38. 外观数列</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>初始字符串从 “1”开始</li>
<li>循环 n 次，找出连续相同的字符串;</li>
<li>然后每次将他们串起来，每次给 s 重新赋值</li>
<li>找相同的数有多少个的方法：</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 从第 j 个位置开始计算有多少个相同的数</span></span><br><span class="line"><span class="type">int</span> k = j;</span><br><span class="line"><span class="comment">// 首先保证s[k]不会溢出，再保证遍历到的值相等，这时计数+1；</span></span><br><span class="line"><span class="keyword">while</span>(k &lt; s.<span class="built_in">size</span>() &amp;&amp; s[j] == s[k]) k++;</span><br><span class="line"><span class="comment">// 最后计算有几个重复的数</span></span><br><span class="line">n = k - j;</span><br></pre></td></tr></table></figure>

<p>代码：</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">string <span class="title">countAndSay</span><span class="params">(<span class="type">int</span> n)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">if</span>(n &lt;= <span class="number">1</span>) <span class="keyword">return</span> <span class="built_in">to_string</span>(n);</span><br><span class="line">        string s = <span class="string">&quot;1&quot;</span>;</span><br><span class="line">        <span class="comment">// 由于第一次默认是1，已经输出过来，所以 i &lt; n-1</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n <span class="number">-1</span>; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            string ns;</span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">0</span>; j &lt; s.<span class="built_in">size</span>(); j++)</span><br><span class="line">            &#123;   </span><br><span class="line">                <span class="comment">// 从第 j 个位置开始计算有多少个相同的数</span></span><br><span class="line">                <span class="type">int</span> k = j;</span><br><span class="line">                <span class="comment">// 首先保证s[k]不会溢出，再保证遍历到的值相等，这时计数+1；</span></span><br><span class="line">                <span class="keyword">while</span>(k &lt; s.<span class="built_in">size</span>() &amp;&amp; s[j] == s[k]) k++;</span><br><span class="line">                <span class="comment">// 把他们串起来</span></span><br><span class="line">                ns += <span class="built_in">to_string</span>(k-j) + s[j];</span><br><span class="line">                <span class="comment">// 更新到第k-1个数</span></span><br><span class="line">                j = k - <span class="number">1</span>;</span><br><span class="line">            &#125;</span><br><span class="line"></span><br><span class="line">            s = ns;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> s; </span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="49-字母异位词分组"><a href="#49-字母异位词分组" class="headerlink" title="49. 字母异位词分组"></a>49. 字母异位词分组</h4><ul>
<li>程序</li>
</ul>
<ol>
<li>创建一个哈希表：先将字符串按照字母顺序排序，直接用sort函数；</li>
<li>此时排序后相同的字符串归为一类，以排序后的字符串为键，未排序前的字符串为值，添加进哈希表中；</li>
<li>最后将哈希表的元素取出来即可</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    vector&lt;vector&lt;string&gt;&gt; <span class="built_in">groupAnagrams</span>(vector&lt;string&gt;&amp; strs) &#123;</span><br><span class="line">        </span><br><span class="line">        <span class="comment">// 定义一个哈希表</span></span><br><span class="line">        unordered_map&lt;string, vector&lt;string&gt;&gt; hash;</span><br><span class="line">        <span class="comment">// 遍历元素</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> it : strs)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 备份，否则排序完字符串就发生改变了</span></span><br><span class="line">            string s = it;</span><br><span class="line">            <span class="comment">// 排序</span></span><br><span class="line">            <span class="built_in">sort</span>(s.<span class="built_in">begin</span>(), s.<span class="built_in">end</span>());</span><br><span class="line">            hash[s].<span class="built_in">push_back</span>(it);</span><br><span class="line">        &#125;</span><br><span class="line">        vector&lt;vector&lt;string&gt;&gt; res;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> item : hash)</span><br><span class="line">        &#123;</span><br><span class="line">            res.<span class="built_in">push_back</span>(item.second);</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="191-反转字符串中的单词"><a href="#191-反转字符串中的单词" class="headerlink" title="191. 反转字符串中的单词"></a>191. 反转字符串中的单词</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><p><strong>方法一：</strong>搜寻完整的单词，将其存到栈中；</p>
</li>
<li><p>逐个取出栈中的元素</p>
</li>
<li><p><strong>方法二：</strong>翻转每个单词；再反转整个字符串；</p>
</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">string <span class="title">reverseWords</span><span class="params">(string s)</span> </span>&#123;</span><br><span class="line">        </span><br><span class="line">        stack&lt;string&gt; stacks;</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; s.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 遍历到不是空格的</span></span><br><span class="line">            <span class="keyword">while</span>(i &lt; s.<span class="built_in">size</span>() &amp;&amp; s[i] == <span class="string">&#x27; &#x27;</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                i++;</span><br><span class="line">            &#125; </span><br><span class="line">            <span class="keyword">if</span>(i == s.<span class="built_in">size</span>()) <span class="keyword">break</span>;</span><br><span class="line">            <span class="type">int</span> j = i;</span><br><span class="line">            <span class="keyword">while</span>(j &lt; s.<span class="built_in">size</span>() &amp;&amp; s[j] != <span class="string">&#x27; &#x27;</span>)</span><br><span class="line">            &#123;</span><br><span class="line">                j++;</span><br><span class="line">            &#125; </span><br><span class="line">            stacks.<span class="built_in">push</span>(s.<span class="built_in">substr</span>(i, j-i));</span><br><span class="line">            <span class="comment">// 再从当前位置继续判断下去</span></span><br><span class="line">            i = j;</span><br><span class="line">        &#125;</span><br><span class="line">        string strs;</span><br><span class="line">        <span class="keyword">while</span>(!stacks.<span class="built_in">empty</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            strs +=  stacks.<span class="built_in">top</span>() + <span class="string">&#x27; &#x27;</span>;</span><br><span class="line">            stacks.<span class="built_in">pop</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        strs.<span class="built_in">erase</span>(strs.<span class="built_in">end</span>() <span class="number">-1</span>);</span><br><span class="line">        <span class="keyword">return</span> strs;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="165-比较版本号"><a href="#165-比较版本号" class="headerlink" title="165. 比较版本号"></a>165. 比较版本号</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>分别从version1和version2中取出完整的整数</li>
<li>将这些整数进行除0操作：将数字前的0删掉，保留0后面的数字；直接用atoi() 函数将取出的字符串数字进行比较大小即可；</li>
<li>注意需要添加一个 k1 &#x3D;&#x3D; i , k2 &#x3D;&#x3D; j 的判断，如果一个字符串没有另一个长，则后面的数用 0.0.0… 补上</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">compareVersion</span><span class="params">(string version1, string version2)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="type">int</span> i = <span class="number">0</span>, j = <span class="number">0</span>;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">while</span>(i &lt; version1.<span class="built_in">size</span>() || j &lt; version2.<span class="built_in">size</span>())</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 拿出整数字符串</span></span><br><span class="line">            <span class="type">int</span> k1 = i, k2 = j;</span><br><span class="line">            <span class="keyword">while</span>(k1 &lt; version1.<span class="built_in">size</span>() &amp;&amp; version1[k1] != <span class="string">&#x27;.&#x27;</span>) k1++;</span><br><span class="line">            <span class="keyword">while</span>(k2 &lt; version2.<span class="built_in">size</span>() &amp;&amp; version2[k2] != <span class="string">&#x27;.&#x27;</span>) k2++;</span><br><span class="line">            <span class="comment">// 将整数字符串取出并转为数字</span></span><br><span class="line">            <span class="type">int</span> x = k1 == i ? <span class="number">0</span> : <span class="built_in">atoi</span>(version1.<span class="built_in">substr</span>(i, k1 - i).<span class="built_in">c_str</span>());</span><br><span class="line">            <span class="type">int</span> y = k2 == j ? <span class="number">0</span> : <span class="built_in">atoi</span>(version2.<span class="built_in">substr</span>(j, k2 - j).<span class="built_in">c_str</span>());</span><br><span class="line">            <span class="comment">// 比较大小</span></span><br><span class="line">            <span class="keyword">if</span>(x &gt; y) <span class="keyword">return</span> <span class="number">1</span>;</span><br><span class="line">            <span class="keyword">if</span>(x &lt; y) <span class="keyword">return</span> <span class="number">-1</span>;</span><br><span class="line">            i = k1 + <span class="number">1</span>;</span><br><span class="line">            j = k2 + <span class="number">1</span>;</span><br><span class="line"></span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="929-独特的电子邮件地址"><a href="#929-独特的电子邮件地址" class="headerlink" title="929. 独特的电子邮件地址"></a>929. 独特的电子邮件地址</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>将邮箱分成用户名和域名，将用户名的’.’ 删去，将’+’号以后的全部删除</li>
<li>将处理过的用户名的域名拼接起来作为哈希表的键，原来的名称为值，存进表中；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">numUniqueEmails</span><span class="params">(vector&lt;string&gt;&amp; emails)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        unordered_map&lt;string, vector&lt;string&gt;&gt; hash;</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> email : emails)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 将用户名和域名分开</span></span><br><span class="line">            <span class="type">int</span> at = email.<span class="built_in">find</span>(<span class="string">&#x27;@&#x27;</span>);</span><br><span class="line">            <span class="comment">// 用户名</span></span><br><span class="line">            string name;</span><br><span class="line">            <span class="comment">// 取出用户名中的 &#x27;.&#x27; 和 &#x27;+&#x27;</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">auto</span> c : email.<span class="built_in">substr</span>(<span class="number">0</span>, at))</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">if</span>(c == <span class="string">&#x27;.&#x27;</span>) <span class="keyword">continue</span>;</span><br><span class="line">                <span class="keyword">else</span> <span class="keyword">if</span>(c == <span class="string">&#x27;+&#x27;</span>) <span class="keyword">break</span>;</span><br><span class="line">                <span class="keyword">else</span> name += c;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 拿到域名</span></span><br><span class="line">            string domain = email.<span class="built_in">substr</span>(at + <span class="number">1</span>);</span><br><span class="line">            hash[name+<span class="string">&quot;@&quot;</span>+domain].<span class="built_in">push_back</span>(email);    </span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> hash.<span class="built_in">size</span>();</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="3-无重复字符的最长子串"><a href="#3-无重复字符的最长子串" class="headerlink" title="3. 无重复字符的最长子串"></a>3. 无重复字符的最长子串</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>遍历每个数</li>
<li>利用右指针在不溢出的情况下，去找跟他相同的数；</li>
<li>将遍历到数都存进set，找到每个不等于set里的所有元素的元素；</li>
<li>如果长度比她长就更新这个字串；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">lengthOfLongestSubstring</span><span class="params">(string s)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(s.<span class="built_in">size</span>() &lt;= <span class="number">1</span>) <span class="keyword">return</span> s.<span class="built_in">size</span>();</span><br><span class="line">        string res;</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; s.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> right = i;</span><br><span class="line">            std::set&lt;<span class="type">char</span>&gt; sets;</span><br><span class="line">            <span class="keyword">while</span>(right &lt; s.<span class="built_in">size</span>())</span><br><span class="line">            &#123;   </span><br><span class="line">                <span class="comment">// 判断该元素是否出现过</span></span><br><span class="line">                <span class="keyword">if</span>(sets.<span class="built_in">count</span>(s[right]) != <span class="number">0</span>) <span class="keyword">break</span>;</span><br><span class="line">                <span class="comment">// 将元素添加进容器中</span></span><br><span class="line">                sets.<span class="built_in">insert</span>(s[right]);</span><br><span class="line">                right ++;</span><br><span class="line">            &#125; </span><br><span class="line">            <span class="keyword">if</span>(res.<span class="built_in">size</span>() &lt;  right - i) res = s.<span class="built_in">substr</span>(i, right - i);</span><br><span class="line">            </span><br><span class="line">            <span class="comment">// 清空set;</span></span><br><span class="line">            sets.<span class="built_in">clear</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res.<span class="built_in">size</span>();</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h4 id="5-最长回文串"><a href="#5-最长回文串" class="headerlink" title="5. 最长回文串"></a>5. 最长回文串</h4><ul>
<li><strong>思路</strong></li>
</ul>
<p>回文串是对称的；回文串长度有奇、偶数的情况：奇数-中心点是一个数；偶数-中心点是两个数；</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><p>用两个指针，去找对称的元素；</p>
</li>
<li><p>指针在不溢出的情况下，找到满足 s[left] &#x3D;&#x3D; s[right] 的元素；</p>
</li>
<li><p>遍历每个元素，然后以每个元素为中心点去找它两边对称的元素；</p>
</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">string <span class="title">longestPalindrome</span><span class="params">(string s)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        string res;</span><br><span class="line">        <span class="comment">// 遍历每一个数，找它左右两边对称的数</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; s.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 当中心点是一个数的时候</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> l = i, r = i; l&gt;= <span class="number">0</span> &amp;&amp; r &lt; s.<span class="built_in">size</span>() &amp;&amp; s[l] == s[r]; l--, r++)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">if</span>(res.<span class="built_in">size</span>() &lt; r - l + <span class="number">1</span>) res = s.<span class="built_in">substr</span>(l, r - l + <span class="number">1</span>);</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 当中心点是两个对称的数的时候</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> l = i, r = i + <span class="number">1</span>; l &gt;= <span class="number">0</span> &amp;&amp; r &lt; s.<span class="built_in">size</span>() &amp;&amp; s[l] == s[r]; l--, r++)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">if</span>(res.<span class="built_in">size</span>() &lt; r - l + <span class="number">1</span>) res = s.<span class="built_in">substr</span>(l , r - l + <span class="number">1</span>);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line">    &#125;</span><br></pre></td></tr></table></figure>

<h4 id="208-实现-trie-前缀树"><a href="#208-实现-trie-前缀树" class="headerlink" title="208. 实现 trie 前缀树"></a>208. 实现 trie 前缀树</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>因为是树形数据结构，所以必须有节点；创建节点结构体；</li>
<li>结构体包括：1个单词结尾标志位end；26个子节点指针；初始化指针；</li>
<li>insert：遍历待插入的单词，从头节点开始搜索，如果有，则跳到其子节点，如果没有就创建新的子节点，然后移动到该节点；</li>
<li>search：遍历待搜索的单词，从头节点开始搜索；如果有，则跳到子节点；如果没有，则直接返回 false；遍历结束，返回 单词结尾标志位 end;</li>
<li>startsWith：遍历搜索的前缀，从头节点开始搜索，如果有，则跳到子节点；如果没有，则直接返回fasle；遍历结束，返回 true;</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br><span class="line">63</span><br><span class="line">64</span><br><span class="line">65</span><br><span class="line">66</span><br><span class="line">67</span><br><span class="line">68</span><br><span class="line">69</span><br><span class="line">70</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Trie</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="keyword">struct</span> <span class="title class_">Node</span></span><br><span class="line">    &#123;</span><br><span class="line">        <span class="comment">// 值域：是否结尾</span></span><br><span class="line">        <span class="type">bool</span> is_end;</span><br><span class="line">        <span class="comment">// 指针域：有每个字母26个子节点</span></span><br><span class="line">        Node *son[<span class="number">26</span>];  </span><br><span class="line">        <span class="comment">// 构造函数</span></span><br><span class="line">        <span class="built_in">Node</span>()</span><br><span class="line">        &#123;</span><br><span class="line">            is_end = <span class="literal">false</span>;</span><br><span class="line">            <span class="comment">// 没有赋值的空指针一定要设为 NULL</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; <span class="number">26</span>; i++) son[i] = <span class="literal">NULL</span>;</span><br><span class="line"></span><br><span class="line">        &#125;  </span><br><span class="line">    &#125;;</span><br><span class="line">    <span class="comment">// 定义一个根节点</span></span><br><span class="line">    Node *root;</span><br><span class="line"></span><br><span class="line">    <span class="built_in">Trie</span>() &#123;</span><br><span class="line">        root = <span class="keyword">new</span> <span class="built_in">Node</span>();</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">insert</span><span class="params">(string word)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 从根节点开始</span></span><br><span class="line">        <span class="keyword">auto</span> p = root;</span><br><span class="line">        <span class="comment">// 遍历插入单词的每一个字母</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> c  : word)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 编号</span></span><br><span class="line">            <span class="type">int</span> u = c - <span class="string">&#x27;a&#x27;</span>;</span><br><span class="line">            <span class="comment">// 如果子节点中没有，则创建一个新的子节点</span></span><br><span class="line">            <span class="keyword">if</span>(p-&gt;son[u] == <span class="literal">NULL</span>) p-&gt;son[u] = <span class="keyword">new</span> <span class="built_in">Node</span>();</span><br><span class="line">            <span class="comment">// p 再移动到子节点的位置</span></span><br><span class="line">            p = p-&gt;son[u];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 有以 p 为结尾的单词</span></span><br><span class="line">        p-&gt;is_end = <span class="literal">true</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">search</span><span class="params">(string word)</span></span>&#123;</span><br><span class="line">        <span class="comment">// 从根节点开始</span></span><br><span class="line">        <span class="keyword">auto</span> p = root;    </span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> c : word)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 编号</span></span><br><span class="line">            <span class="type">int</span> u = c - <span class="string">&#x27;a&#x27;</span>;</span><br><span class="line">            <span class="comment">// 判断子节点中是否有，没有就返回false</span></span><br><span class="line">            <span class="keyword">if</span>(p-&gt;son[u] == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">            <span class="comment">// 有的话就移动到下一个子节点</span></span><br><span class="line">            p = p-&gt;son[u];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 判断是否有这个单词</span></span><br><span class="line">        <span class="keyword">return</span> p-&gt;is_end;</span><br><span class="line">    &#125;</span><br><span class="line">    </span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">startsWith</span><span class="params">(string prefix)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">auto</span> p = root;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> c : prefix)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="type">int</span> u = c - <span class="string">&#x27;a&#x27;</span>;</span><br><span class="line">            <span class="comment">// 判断当前节点是否有这个字符，没有的话就返回false</span></span><br><span class="line">            <span class="keyword">if</span>(p-&gt;son[u] == <span class="literal">NULL</span>) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">            <span class="comment">// 有的话就移动到下一个子节点</span></span><br><span class="line">            p = p-&gt;son[u];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="8-字符串转换整数-atoi"><a href="#8-字符串转换整数-atoi" class="headerlink" title="8. 字符串转换整数 - atoi"></a>8. 字符串转换整数 - atoi</h4><ul>
<li>程序</li>
</ul>
<ol>
<li>首先去除空格字符</li>
<li>拿出正负号</li>
<li>取出完整的整数部分</li>
<li>对整数部分进行特殊处理：1. 如果第一个数不为0，那么就只取11位，否则会溢出 int 型的空间；否则就直接取</li>
<li>将字符串转换为10进制的数</li>
<li>判断是否在 [-2147483648, 2147483647] 范围内；</li>
<li>结束</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">myAtoi</span><span class="params">(string s)</span> </span>&#123;</span><br><span class="line">        string num;</span><br><span class="line">        <span class="type">long</span> <span class="type">long</span> <span class="type">int</span> ans = <span class="number">0</span>;</span><br><span class="line">        <span class="type">char</span> flag;</span><br><span class="line">        <span class="comment">// 去除空格</span></span><br><span class="line">        <span class="type">int</span> k = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">while</span>(k &lt; s.<span class="built_in">size</span>() &amp;&amp; s[k] == <span class="string">&#x27; &#x27;</span>) k++;</span><br><span class="line">        <span class="comment">// 找到正负号</span></span><br><span class="line">        <span class="keyword">if</span>(s[k] == <span class="string">&#x27;-&#x27;</span> || s[k] == <span class="string">&#x27;+&#x27;</span> ) flag = s[k++];</span><br><span class="line">        <span class="comment">// 记录当前第一个数字的位置</span></span><br><span class="line">        <span class="type">int</span> w = k;</span><br><span class="line">        <span class="comment">// 找到到数字的整体部分</span></span><br><span class="line">        <span class="keyword">while</span>(w &lt; s.<span class="built_in">size</span>() &amp;&amp; (<span class="number">0</span> &lt;= (s[w] - <span class="string">&#x27;0&#x27;</span>)) &amp;&amp; ((s[w] - <span class="string">&#x27;0&#x27;</span>) &lt;= <span class="number">9</span>)) w++;</span><br><span class="line">        <span class="comment">// 取出数字字符串</span></span><br><span class="line">        num = s.<span class="built_in">substr</span>(k, w - k);</span><br><span class="line">        <span class="keyword">if</span>(num[<span class="number">0</span>] != <span class="string">&#x27;0&#x27;</span>) num = num.<span class="built_in">substr</span>(<span class="number">0</span>, <span class="number">11</span>);</span><br><span class="line">        <span class="comment">// 遍历纯数字的字符串</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; num.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            ans += (num[i] - <span class="string">&#x27;0&#x27;</span>) * <span class="built_in">pow</span>(<span class="number">10</span>, num.<span class="built_in">size</span>() - i - <span class="number">1</span>);</span><br><span class="line">        &#125;</span><br><span class="line">        </span><br><span class="line">        <span class="comment">// 取正负</span></span><br><span class="line">        <span class="keyword">if</span>(flag == <span class="string">&#x27;-&#x27;</span>) ans = -ans;</span><br><span class="line">        <span class="comment">// 截断数字</span></span><br><span class="line">        ans = ans &gt;= <span class="number">2147483647</span> ? <span class="number">2147483647</span> : ans;</span><br><span class="line">        ans = ans &lt;= <span class="number">-2147483648</span> ? <span class="number">-2147483648</span> : ans;</span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line"></span><br><span class="line">       </span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>



<h3 id="双指针（2）"><a href="#双指针（2）" class="headerlink" title="双指针（2）"></a>双指针（2）</h3><h4 id="167-两数之和-ii-输入有序数组"><a href="#167-两数之和-ii-输入有序数组" class="headerlink" title="167. 两数之和-ii-输入有序数组"></a>167. 两数之和-ii-输入有序数组</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>双指针法：</strong>定义头尾指针 l, r; 分别移动头尾指针；如何移动是需要思考的问题 ？；满足 nums[l] + nums[r-1] &gt;&#x3D; target 就移动右指针，不满足就移动左指针；左右两边根据条件移动，只需要遍历一次即可；</li>
<li>固定头指针，判断如果 nums[l] + nums[r - 1] &gt;&#x3D; target，那么就移动右指针 r–；这里使用 i-1的原因是当条件 nums[l] + nums[r - 1] &gt;&#x3D; target 成立，那么此时会执行 r– 操作，r– 后下一次判断条件 nums[l] + nums[r - 1] &gt;&#x3D; target 将不在成立；此时输出的 r 刚好是最后一个满足 nums[l] + nums[r] &gt;&#x3D; target 的数；</li>
<li>l, r 有单调性，r 往左走，那么 l 必是往右走，不可能往左走；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;<span class="type">int</span>&gt; <span class="title">twoSum</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; numbers, <span class="type">int</span> target)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> l = <span class="number">0</span>, r = numbers.<span class="built_in">size</span>() - <span class="number">1</span>; l &lt; numbers.<span class="built_in">size</span>(); l++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 找到满足条件 nums[l] + nums[r] &gt;= target的值，就缩小右指针的范围</span></span><br><span class="line">            <span class="keyword">while</span>(r - <span class="number">1</span> &gt; <span class="number">0</span> &amp;&amp; r - <span class="number">1</span> &gt; l &amp;&amp; numbers[l] + numbers[r <span class="number">-1</span>] &gt;= target) r--;</span><br><span class="line">            <span class="keyword">if</span>(numbers[l] + numbers[r] == target) <span class="keyword">return</span> &#123;l + <span class="number">1</span>, r + <span class="number">1</span>&#125;;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> &#123;<span class="number">-1</span>, <span class="number">-1</span>&#125;;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="26-删除有序数组中的重复项"><a href="#26-删除有序数组中的重复项" class="headerlink" title="26. 删除有序数组中的重复项"></a>26. 删除有序数组中的重复项</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li><strong>双指针法：</strong>慢指针用来保存不同的元素，快指针用来找不同的元素</li>
<li>如果 nums[r] !&#x3D; nums[l] ，则左指针往右移动一格，在将nums[r] 的值赋值给他</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function"><span class="type">int</span> <span class="title">removeDuplicates</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums)</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">        <span class="keyword">if</span>(nums.<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="number">0</span>;</span><br><span class="line">        <span class="type">int</span> l = <span class="number">0</span>;</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> r = <span class="number">0</span>; r &lt; nums.<span class="built_in">size</span>(); r++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">if</span>(nums[l] != nums[r]) nums[++l] = nums[r];</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> l + <span class="number">1</span>;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h3 id="滑动窗口（1）"><a href="#滑动窗口（1）" class="headerlink" title="滑动窗口（1）"></a>滑动窗口（1）</h3><h4 id="76-最小覆盖子串"><a href="#76-最小覆盖子串" class="headerlink" title="76. 最小覆盖子串"></a>76. 最小覆盖子串</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.23.png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<p>1. </p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">string <span class="title">minWindow</span><span class="params">(string s, string t)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 记录 t 中字母的次数</span></span><br><span class="line">        unordered_map&lt;<span class="type">char</span>, <span class="type">int</span>&gt; hash;</span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> c : t) hash[c] ++;</span><br><span class="line">        <span class="comment">// 不同字母的个数</span></span><br><span class="line">        <span class="type">int</span> cnt = hash.<span class="built_in">size</span>();</span><br><span class="line"></span><br><span class="line">        string res;</span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>, j = <span class="number">0</span>, c = <span class="number">0</span>; i &lt; s.<span class="built_in">size</span>(); i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// </span></span><br><span class="line">            <span class="keyword">if</span>(hash[s[i]] == <span class="number">1</span>) c++;</span><br><span class="line">            <span class="comment">//</span></span><br><span class="line">            hash[s[i]] --;</span><br><span class="line">            <span class="keyword">while</span>(hash[s[j]] &lt; <span class="number">0</span>) hash[s[j++]]++;</span><br><span class="line">            <span class="keyword">if</span>(c == cnt)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="keyword">if</span>(res.<span class="built_in">empty</span>() || res.<span class="built_in">size</span>() &gt;  i - j + <span class="number">1</span>) res = s.<span class="built_in">substr</span>(j, i - j + <span class="number">1</span>);</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> res;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h3 id="DFS（深度搜索）-回溯（4）"><a href="#DFS（深度搜索）-回溯（4）" class="headerlink" title="DFS（深度搜索） + 回溯（4）"></a>DFS（深度搜索） + 回溯（4）</h3><h4 id="17-电话号码的字母组合"><a href="#17-电话号码的字母组合" class="headerlink" title="17. 电话号码的字母组合"></a>17. 电话号码的字母组合</h4><ul>
<li><strong>思路</strong></li>
</ul>
<p>从第一个数字的字符开始与空集相组合，得到第一次的组合结果；第二个数字的字符与第一次的组合结果依次组合，得到第二次的组合结果；循环往下….</p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>用一个 string 变量存储数字所对应的字符 string a[8] &#x3D; {“adc”,”def”,……};</li>
<li>创建一个容器用来表示上一次的组合结果，再创建一个用来保存当前组合的结果；初始的组合结果为 “ ”；</li>
<li>每次把遍历到的数字所对应的每个字符与上一次组合的结果相组合；得到当前的组合结果，再保存为上一次结果；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="function">vector&lt;string&gt; <span class="title">letterCombinations</span><span class="params">(string digits)</span> </span>&#123;</span><br><span class="line">        <span class="keyword">if</span>(digits.<span class="built_in">empty</span>()) <span class="keyword">return</span> &#123;&#125;;</span><br><span class="line"></span><br><span class="line">        string chars[<span class="number">8</span>] = &#123;<span class="string">&quot;abc&quot;</span>, <span class="string">&quot;def&quot;</span>, <span class="string">&quot;ghi&quot;</span>, <span class="string">&quot;jkl&quot;</span>, <span class="string">&quot;mno&quot;</span>, <span class="string">&quot;pqrs&quot;</span>, <span class="string">&quot;tuv&quot;</span>, <span class="string">&quot;wxyz&quot;</span>&#125;;</span><br><span class="line">        <span class="comment">// 创建上一次的组合结果变量、当前的组合结果变量</span></span><br><span class="line">        vector&lt;string&gt; pre, now;</span><br><span class="line">        <span class="comment">// 初始化上一次组合结果为 &quot;&quot;，不初始化的话会导致下面与它组合的所有操作为空</span></span><br><span class="line">        pre = &#123;<span class="string">&quot;&quot;</span>&#125;;</span><br><span class="line">        <span class="comment">// 遍历每一个数字</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="keyword">auto</span> n :  digits)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 遍历数字对应的每一个字符</span></span><br><span class="line">            <span class="keyword">for</span>(<span class="keyword">auto</span> c : chars[n - <span class="string">&#x27;2&#x27;</span>])</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 组合上一次结果的每一个元素</span></span><br><span class="line">                <span class="keyword">for</span>(<span class="keyword">auto</span> s : pre)</span><br><span class="line">                &#123;</span><br><span class="line">                    now.<span class="built_in">push_back</span>(s + c);</span><br><span class="line">                &#125;</span><br><span class="line">            &#125;</span><br><span class="line">            <span class="comment">// 保存上一次结果</span></span><br><span class="line">            pre = now;</span><br><span class="line">            <span class="comment">// 清空容器;</span></span><br><span class="line">            now.<span class="built_in">clear</span>();</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="keyword">return</span> pre;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="79-单词搜索"><a href="#79-单词搜索" class="headerlink" title="79. 单词搜索"></a>79. 单词搜索</h4><ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>枚举起点，对每个位置进行 DFS 搜索；</li>
<li>DFS 搜索从起点开始，依次搜索其上下左右位置的点是否等于待搜索字母；</li>
<li>DFS 搜索的递归结束条件：1. 如果当前的元素不等于待搜索的元素，则直接搜索失败返回 false，进行下一个起点的搜索； 2. 如果当前元素等于待搜索的元素，并且已经是最后一个点了，那么搜索成功，直接返回 true，结束程序；</li>
<li>递归的执行函数：先将到达点的字母置为 ‘0’ ；再对上下左右的位置进行下一个字母的搜索，搜索结束；将先前置位的点的字母还原；</li>
</ol>
<p>	</p>
<p><strong>一个取坐标的上下左右技巧</strong></p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br></pre></td><td class="code"><pre><span class="line"><span class="comment">// 上 右 下 左</span></span><br><span class="line"><span class="type">int</span> dx[<span class="number">4</span>] = &#123;<span class="number">-1</span>, <span class="number">0</span>, <span class="number">1</span>, <span class="number">0</span>&#125;, dy[<span class="number">4</span>] = &#123;<span class="number">0</span>, <span class="number">1</span>, <span class="number">0</span>, <span class="number">-1</span>&#125;;</span><br><span class="line"><span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; <span class="number">4</span>; i ++)</span><br><span class="line">&#123;</span><br><span class="line">    <span class="comment">// 依次计算 上 右 下 左 的坐标</span></span><br><span class="line">    <span class="type">int</span> a = x + dx[i], b = y + dy[i];</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

<p>代码：</p>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br><span class="line">52</span><br><span class="line">53</span><br><span class="line">54</span><br><span class="line">55</span><br><span class="line">56</span><br><span class="line">57</span><br><span class="line">58</span><br><span class="line">59</span><br><span class="line">60</span><br><span class="line">61</span><br><span class="line">62</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"></span><br><span class="line">    <span class="type">int</span> n, m;</span><br><span class="line">    <span class="comment">// 上 右 下 左</span></span><br><span class="line">    <span class="type">int</span> dx[<span class="number">4</span>] = &#123;<span class="number">-1</span>, <span class="number">0</span>, <span class="number">1</span>, <span class="number">0</span>&#125;, dy[<span class="number">4</span>] = &#123;<span class="number">0</span>, <span class="number">1</span>, <span class="number">0</span>, <span class="number">-1</span>&#125;;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">exist</span><span class="params">(vector&lt;vector&lt;<span class="type">char</span>&gt;&gt;&amp; board, string word)</span> </span>&#123;</span><br><span class="line">        <span class="comment">// 1. 判空</span></span><br><span class="line">        <span class="keyword">if</span>(board.<span class="built_in">empty</span>() || board[<span class="number">0</span>].<span class="built_in">empty</span>()) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">        <span class="comment">// 行</span></span><br><span class="line">        n = board.<span class="built_in">size</span>(); </span><br><span class="line">        <span class="comment">// 列</span></span><br><span class="line">        m = board[<span class="number">0</span>].<span class="built_in">size</span>(); </span><br><span class="line">        <span class="comment">// 枚举起点</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i ++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="keyword">for</span>(<span class="type">int</span> j = <span class="number">0</span>; j &lt; m; j ++)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 如果有一个点搜索成功，则返回true</span></span><br><span class="line">                <span class="keyword">if</span>(<span class="built_in">DFS</span>(board, i, j, word, <span class="number">0</span>)) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">                <span class="comment">// 搜索失败，则枚举下一个起点</span></span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 否则，返回false</span></span><br><span class="line">        <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line">    <span class="comment">/**</span></span><br><span class="line"><span class="comment">     * 深度搜索 DFS</span></span><br><span class="line"><span class="comment">     * @param x 横坐标</span></span><br><span class="line"><span class="comment">     * @param y 纵坐标</span></span><br><span class="line"><span class="comment">     * @param u 当前待搜索字母的下标</span></span><br><span class="line"><span class="comment">    */</span></span><br><span class="line">    <span class="function"><span class="type">bool</span> <span class="title">DFS</span><span class="params">(vector&lt;vector&lt;<span class="type">char</span>&gt;&gt;&amp; board, <span class="type">int</span> x, <span class="type">int</span> y, string&amp; word, <span class="type">int</span> u)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 如果当前元素不等于待搜索元素，则停止深度搜索</span></span><br><span class="line">        <span class="keyword">if</span>(board[x][y] != word[u]) <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">        <span class="comment">// 如果是待搜索的字母，且已经搜索到最后一个字母了，则返回 true</span></span><br><span class="line">        <span class="keyword">if</span>(u == word.<span class="built_in">size</span>() - <span class="number">1</span>) <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">		<span class="comment">// 赋值标志位，保证其不会往回搜索</span></span><br><span class="line">        board[x][y] = <span class="string">&#x27;.&#x27;</span>;</span><br><span class="line">        <span class="comment">// 进行搜索下一个字母，位于当前的上下左右的位置</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; <span class="number">4</span>; i ++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 计算上下左右坐标位置</span></span><br><span class="line">            <span class="type">int</span> a = x + dx[i], b = y + dy[i];</span><br><span class="line">            <span class="comment">// 如果在范围内</span></span><br><span class="line">            <span class="keyword">if</span>(a &gt;= <span class="number">0</span> &amp;&amp; a &lt; n &amp;&amp; b &gt;= <span class="number">0</span> &amp;&amp; b &lt; m)</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 继续搜索四个方向,下一个字母</span></span><br><span class="line">                <span class="keyword">if</span>(<span class="built_in">DFS</span>(board, a, b, word, u + <span class="number">1</span>)) </span><br><span class="line">                    <span class="comment">// 如果其中有一个方向搜索成功，直接返回 true</span></span><br><span class="line">                    <span class="keyword">return</span> <span class="literal">true</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 回溯该点的值，为进行下一次搜索做准备</span></span><br><span class="line">        board[x][y] = word[u];</span><br><span class="line">		<span class="comment">// 搜索失败，搜索下一个起点</span></span><br><span class="line">        <span class="keyword">return</span> <span class="literal">false</span>;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>

<h4 id="46-全排列"><a href="#46-全排列" class="headerlink" title="46. 全排列"></a>46. 全排列</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.25(15.06).png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>枚举每个位置上放哪个数；</li>
<li>创建一个容器用来存放结果，创建一个容器用来存放临时组合结果，创建一个个容器用来判断当前的数字是否可以用来放置；</li>
<li>从第一个位置开始 DFS 搜索；</li>
<li>递归终止条件：如果当前的位置是最后一个位置，那么将一个临时的结果保存后，再结束 return；</li>
<li>递归的执行函数：遍历当前位置可以填的数，从第一个数开始，判断当前的数字是否可以用，如果还没用 false 的话就可以取用（判断 st[i] 是否为 false）；可以用的话，先将该数字锁死置为 true 后（这个数下次将不再能被调用），再将该数字存入临时结果，并执行下一层递归；递归到最后一个位置结束后，将数字锁打开，清除临时结果里的那个添加的值；</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line">    <span class="type">int</span> n;</span><br><span class="line">    <span class="comment">// 当前分支可以用的数字有哪些</span></span><br><span class="line">    vector&lt;<span class="type">bool</span>&gt; st;</span><br><span class="line">    <span class="comment">// 存答案</span></span><br><span class="line">    vector&lt;vector&lt;<span class="type">int</span>&gt;&gt; ans;</span><br><span class="line">    <span class="comment">// 存临时组合结果</span></span><br><span class="line">    vector&lt;<span class="type">int</span>&gt; path;</span><br><span class="line"></span><br><span class="line">    vector&lt;vector&lt;<span class="type">int</span>&gt;&gt; <span class="built_in">permute</span>(vector&lt;<span class="type">int</span>&gt;&amp; nums) &#123;</span><br><span class="line"></span><br><span class="line">        n = nums.<span class="built_in">size</span>();</span><br><span class="line">		<span class="comment">// 初始化后，全为0，也就是 false;</span></span><br><span class="line">        st = <span class="built_in">vector</span>&lt;<span class="type">bool</span>&gt;(n);</span><br><span class="line">		<span class="comment">// 从第一个位置开始搜索</span></span><br><span class="line">        <span class="built_in">DFS</span>(nums, <span class="number">0</span>);</span><br><span class="line">        </span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">DFS</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums, <span class="type">int</span> u)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 递归终止条件：如果遍历完所有方案</span></span><br><span class="line">        <span class="keyword">if</span>( u == n)</span><br><span class="line">        &#123;</span><br><span class="line">            ans.<span class="built_in">push_back</span>(path);</span><br><span class="line">            <span class="keyword">return</span>;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 否则遍历当前的格子可以填的所有数</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = <span class="number">0</span>; i &lt; n; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 当为 false 时，说明数字没被锁死，可以用</span></span><br><span class="line">            <span class="keyword">if</span>(!s[i])</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 先将数字锁死</span></span><br><span class="line">                s[i] = <span class="literal">true</span>;</span><br><span class="line">                <span class="comment">// 再存入临时结果</span></span><br><span class="line">                path.<span class="built_in">push_back</span>(nums[i]);</span><br><span class="line">                <span class="comment">// 进行下一个位置的搜索</span></span><br><span class="line">                <span class="built_in">DFS</span>(nums, u + <span class="number">1</span>);</span><br><span class="line">                <span class="comment">// 恢复现场</span></span><br><span class="line">                path.<span class="built_in">pop_back</span>();</span><br><span class="line">                <span class="comment">// 解锁，以便下一次可以调用这个数字</span></span><br><span class="line">                s[i] = <span class="literal">false</span>;</span><br><span class="line"></span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br><span class="line"></span><br></pre></td></tr></table></figure>

<h4 id="47-全排列-Ⅱ"><a href="#47-全排列-Ⅱ" class="headerlink" title="47. 全排列 Ⅱ"></a>47. 全排列 Ⅱ</h4><p><img src="https://gitee.com/fan-jianxuan/pic-go/raw/master/images/2023.7.25(15.47).png"></p>
<ul>
<li><strong>程序</strong></li>
</ul>
<ol>
<li>枚举每个数放到那个位置上；</li>
<li>创建一个容易用来存储结果，创建一个容器用来存储临时组合结果，创建一个容器用来判断该位置是否可以放置数字；</li>
<li>从第一个数开始递归；</li>
<li>递归的终止条件：如果已经到最后一个数字，则将临时结果存到结果数组中，结束搜索 return;</li>
<li>递归的执行函数：</li>
</ol>
<figure class="highlight c++"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br><span class="line">48</span><br><span class="line">49</span><br><span class="line">50</span><br><span class="line">51</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">class</span> <span class="title class_">Solution</span> &#123;</span><br><span class="line"><span class="keyword">public</span>:</span><br><span class="line"></span><br><span class="line">    <span class="type">int</span> n;</span><br><span class="line">    vector&lt;vector&lt;<span class="type">int</span>&gt;&gt; ans;</span><br><span class="line">    vector&lt;<span class="type">int</span>&gt; path;</span><br><span class="line">    vector&lt;<span class="type">bool</span>&gt; st;</span><br><span class="line"></span><br><span class="line">    vector&lt;vector&lt;<span class="type">int</span>&gt;&gt; <span class="built_in">permuteUnique</span>(vector&lt;<span class="type">int</span>&gt;&amp; nums) &#123;</span><br><span class="line">        </span><br><span class="line">        n = nums.<span class="built_in">size</span>();</span><br><span class="line">        <span class="comment">// 初始化位置锁</span></span><br><span class="line">        st = <span class="built_in">vector</span>&lt;<span class="type">bool</span>&gt;(n);</span><br><span class="line">        <span class="comment">// 初始化临时结果容器</span></span><br><span class="line">        path = <span class="built_in">vector</span>&lt;<span class="type">int</span>&gt;(n);</span><br><span class="line">        <span class="comment">// 对数组进行排序，保证相同的数凑到一起</span></span><br><span class="line">        <span class="built_in">sort</span>(nums.<span class="built_in">begin</span>(), nums.<span class="built_in">end</span>());</span><br><span class="line">        <span class="comment">// 从第一个数字开始搜索</span></span><br><span class="line">        <span class="built_in">DFS</span>(nums, <span class="number">0</span>, <span class="number">0</span>);</span><br><span class="line"></span><br><span class="line">        <span class="keyword">return</span> ans;</span><br><span class="line">    &#125;</span><br><span class="line"></span><br><span class="line">    <span class="function"><span class="type">void</span> <span class="title">DFS</span><span class="params">(vector&lt;<span class="type">int</span>&gt;&amp; nums, <span class="type">int</span> u, <span class="type">int</span> start)</span></span></span><br><span class="line"><span class="function">    </span>&#123;</span><br><span class="line">        <span class="comment">// 如果位置已经全部搜索完</span></span><br><span class="line">        <span class="keyword">if</span>(u == n) </span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 将本次搜索的临时结果存到结果数组中</span></span><br><span class="line">            ans.<span class="built_in">push_back</span>(path);</span><br><span class="line">            <span class="keyword">return</span> ;</span><br><span class="line">        &#125;</span><br><span class="line">        <span class="comment">// 否则，进行下一个位置的搜索</span></span><br><span class="line">        <span class="keyword">for</span>(<span class="type">int</span> i = start; i &lt; n; i++)</span><br><span class="line">        &#123;</span><br><span class="line">            <span class="comment">// 如果该位置可以放置数字</span></span><br><span class="line">            <span class="keyword">if</span>(!st[i])</span><br><span class="line">            &#123;</span><br><span class="line">                <span class="comment">// 锁死该位置</span></span><br><span class="line">                st[i] = <span class="literal">true</span>;</span><br><span class="line">                <span class="comment">// 将数组赋值给临时数组</span></span><br><span class="line">                path[i] = nums[u];</span><br><span class="line">                <span class="comment">// 搜索下一个位置，如果下一个数字的值和当前相等，那就从下一个位置开始；否则，还是从 0 开始放；</span></span><br><span class="line">                <span class="built_in">DFS</span>(nums, u + <span class="number">1</span>, u + <span class="number">1</span> &lt;  n &amp;&amp; nums[u + <span class="number">1</span>] == nums[u] ? i + <span class="number">1</span> : <span class="number">0</span>);</span><br><span class="line">                <span class="comment">// 回溯</span></span><br><span class="line">                st[i] = <span class="literal">false</span>;</span><br><span class="line">            &#125;</span><br><span class="line">        &#125;</span><br><span class="line"></span><br><span class="line">    &#125;</span><br><span class="line">&#125;;</span><br></pre></td></tr></table></figure>


    

    

    

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                    <ol class="toc"><li class="toc-item toc-level-1"><a class="toc-link" href="#%E5%88%B7%E9%A2%98%E8%BF%9B%E5%BA%A6%EF%BC%8866%EF%BC%89"><span class="toc-number">1.</span> <span class="toc-text">刷题进度（66）</span></a></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E5%8D%81%E5%A4%A7%E6%8E%92%E5%BA%8F%E7%AE%97%E6%B3%95"><span class="toc-number">2.</span> <span class="toc-text">十大排序算法</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#1-%E9%80%89%E6%8B%A9%E6%8E%92%E5%BA%8F%EF%BC%88%E4%BA%A4%E6%8D%A2%EF%BC%89"><span class="toc-number">2.0.0.1.</span> <span class="toc-text">1. 选择排序（交换）</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#2-%E6%8F%92%E5%85%A5%E6%8E%92%E5%BA%8F%EF%BC%88%E6%AF%94%E8%BE%83%EF%BC%89"><span class="toc-number">2.0.0.2.</span> <span class="toc-text">2. 插入排序（比较）</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#3-%E5%86%92%E6%B3%A1%E6%8E%92%E5%BA%8F%EF%BC%88%E4%BA%A4%E6%8D%A2%EF%BC%89"><span class="toc-number">2.0.0.3.</span> <span class="toc-text">3. 冒泡排序（交换）</span></a><ol class="toc-child"><li class="toc-item toc-level-5"><a class="toc-link" href="#3-1-%E9%9D%9E%E4%BC%98%E5%8C%96%E7%89%88%E6%9C%AC"><span class="toc-number">2.0.0.3.1.</span> <span class="toc-text">3.1 非优化版本</span></a></li><li class="toc-item toc-level-5"><a class="toc-link" href="#3-2-%E4%BC%98%E5%8C%96%E7%89%88%E6%9C%AC"><span class="toc-number">2.0.0.3.2.</span> <span class="toc-text">3.2 优化版本</span></a></li></ol></li><li class="toc-item toc-level-4"><a class="toc-link" href="#4-%E5%B8%8C%E5%B0%94%E6%8E%92%E5%BA%8F"><span class="toc-number">2.0.0.4.</span> <span class="toc-text">4. 希尔排序</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#5-%E5%BD%92%E5%B9%B6%E6%8E%92%E5%BA%8F"><span class="toc-number">2.0.0.5.</span> <span class="toc-text">5. 归并排序</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#6-%E5%BF%AB%E9%80%9F%E6%8E%92%E5%BA%8F"><span class="toc-number">2.0.0.6.</span> <span class="toc-text">6. 快速排序</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#7-%E5%A0%86%E6%8E%92%E5%BA%8F"><span class="toc-number">2.0.0.7.</span> <span class="toc-text">7. 堆排序</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#8-%E8%AE%A1%E6%95%B0%E6%8E%92%E5%BA%8F"><span class="toc-number">2.0.0.8.</span> <span class="toc-text">8. 计数排序</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#9-%E6%A1%B6%E6%8E%92%E5%BA%8F"><span class="toc-number">2.0.0.9.</span> <span class="toc-text">9. 桶排序</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#10-%E5%9F%BA%E6%95%B0%E6%8E%92%E5%BA%8F"><span class="toc-number">2.0.0.10.</span> <span class="toc-text">10. 基数排序</span></a></li></ol></li></ol></li></ol></li><li class="toc-item toc-level-1"><a class="toc-link" href="#%E9%9D%A2%E8%AF%95%E9%A2%98"><span class="toc-number">3.</span> <span class="toc-text">面试题</span></a><ol class="toc-child"><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%95%B0%E7%BB%84%EF%BC%886%EF%BC%89"><span class="toc-number">3.0.1.</span> <span class="toc-text">数组（6）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#1-%E4%B8%A4%E6%95%B0%E4%B9%8B%E5%92%8C"><span class="toc-number">3.0.1.1.</span> <span class="toc-text">1. 两数之和</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#70-%E7%88%AC%E6%A5%BC%E6%A2%AF"><span class="toc-number">3.0.1.2.</span> <span class="toc-text">70. 爬楼梯</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#88-%E5%90%88%E5%B9%B6%E4%B8%A4%E4%B8%AA%E6%95%B0%E7%BB%84"><span class="toc-number">3.0.1.3.</span> <span class="toc-text">88. 合并两个数组</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#283-%E7%A7%BB%E5%8A%A8%E9%9B%B6"><span class="toc-number">3.0.1.4.</span> <span class="toc-text">283. 移动零</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#448-%E6%89%BE%E5%88%B0%E6%89%80%E6%9C%89%E6%95%B0%E7%BB%84%E4%B8%AD%E6%B6%88%E5%A4%B1%E7%9A%84%E6%95%B0%E5%AD%97"><span class="toc-number">3.0.1.5.</span> <span class="toc-text">448. 找到所有数组中消失的数字</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#4-%E5%AF%BB%E6%89%BE%E4%B8%A4%E4%B8%AA%E6%AD%A3%E5%BA%8F%E6%95%B0%E7%BB%84%E7%9A%84%E4%B8%AD%E4%BD%8D%E6%95%B0"><span class="toc-number">3.0.1.6.</span> <span class="toc-text">4. 寻找两个正序数组的中位数</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E9%93%BE%E8%A1%A8%EF%BC%8814%EF%BC%89"><span class="toc-number">3.0.2.</span> <span class="toc-text">链表（14）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#21-%E5%90%88%E5%B9%B6%E4%B8%A4%E4%B8%AA%E6%9C%89%E5%BA%8F%E9%93%BE%E8%A1%A8"><span class="toc-number">3.0.2.1.</span> <span class="toc-text">21. 合并两个有序链表</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#83-%E5%88%A0%E9%99%A4%E6%8E%92%E5%BA%8F%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E9%87%8D%E5%A4%8D%E5%85%83%E7%B4%A0"><span class="toc-number">3.0.2.2.</span> <span class="toc-text">83. 删除排序链表中的重复元素</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#141-%E7%8E%AF%E5%BD%A2%E9%93%BE%E8%A1%A8"><span class="toc-number">3.0.2.3.</span> <span class="toc-text">141. 环形链表</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#142-%E7%8E%AF%E5%BD%A2%E9%93%BE%E8%A1%A8%E2%85%A1"><span class="toc-number">3.0.2.4.</span> <span class="toc-text">142. 环形链表Ⅱ</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#160-%E7%9B%B8%E4%BA%A4%E9%93%BE%E8%A1%A8"><span class="toc-number">3.0.2.5.</span> <span class="toc-text">160. 相交链表</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#206-%E5%8F%8D%E8%BD%AC%E9%93%BE%E8%A1%A8"><span class="toc-number">3.0.2.6.</span> <span class="toc-text">206. 反转链表</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#92-%E5%8F%8D%E8%BD%AC%E9%93%BE%E8%A1%A8-%E2%85%A1"><span class="toc-number">3.0.2.7.</span> <span class="toc-text">92. 反转链表 Ⅱ</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#234-%E5%9B%9E%E6%96%87%E9%93%BE%E8%A1%A8"><span class="toc-number">3.0.2.8.</span> <span class="toc-text">234. 回文链表</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#876-%E9%93%BE%E8%A1%A8%E7%9A%84%E4%B8%AD%E9%97%B4%E8%8A%82%E7%82%B9"><span class="toc-number">3.0.2.9.</span> <span class="toc-text">876. 链表的中间节点</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#19-%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E5%80%92%E6%95%B0%E7%AC%AC-n-%E4%B8%AA%E8%8A%82%E7%82%B9"><span class="toc-number">3.0.2.10.</span> <span class="toc-text">19. 删除链表中的倒数第 n 个节点</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#2-%E4%B8%A4%E6%95%B0%E7%9B%B8%E5%8A%A0"><span class="toc-number">3.0.2.11.</span> <span class="toc-text">2. 两数相加</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#237-%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E8%8A%82%E7%82%B9"><span class="toc-number">3.0.2.12.</span> <span class="toc-text">237. 删除链表中的节点</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#61-%E6%97%8B%E8%BD%AC%E9%93%BE%E8%A1%A8"><span class="toc-number">3.0.2.13.</span> <span class="toc-text">61. 旋转链表</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#24-%E4%B8%A4%E4%B8%A4%E4%BA%A4%E6%8D%A2%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%9A%84%E8%8A%82%E7%82%B9"><span class="toc-number">3.0.2.14.</span> <span class="toc-text">24. 两两交换链表中的节点</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%A0%88%E4%B8%8E%E9%98%9F%E5%88%97%EF%BC%882%EF%BC%89"><span class="toc-number">3.0.3.</span> <span class="toc-text">栈与队列（2）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#232-%E7%94%A8%E6%A0%88%E5%AE%9E%E7%8E%B0%E9%98%9F%E5%88%97"><span class="toc-number">3.0.3.1.</span> <span class="toc-text">232.  用栈实现队列</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#394-%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%A7%A3%E7%A0%81%EF%BC%88%E2%9D%97%E2%9D%97%E2%9D%97%EF%BC%89"><span class="toc-number">3.0.3.2.</span> <span class="toc-text">394. 字符串解码（❗❗❗）</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%A0%91%EF%BC%887%EF%BC%89"><span class="toc-number">3.0.4.</span> <span class="toc-text">树（7）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#94-%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E4%B8%AD%E5%BA%8F%E9%81%8D%E5%8E%86"><span class="toc-number">3.0.4.1.</span> <span class="toc-text">94. 二叉树的中序遍历</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#144-%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E5%89%8D%E5%BA%8F%E9%81%8D%E5%8E%86"><span class="toc-number">3.0.4.2.</span> <span class="toc-text">144. 二叉树的前序遍历</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#145-%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E5%90%8E%E5%BA%8F%E9%81%8D%E5%8E%86"><span class="toc-number">3.0.4.3.</span> <span class="toc-text">145. 二叉树的后序遍历</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#101-%E5%AF%B9%E7%A7%B0%E4%BA%8C%E5%8F%89%E6%A0%91"><span class="toc-number">3.0.4.4.</span> <span class="toc-text">101. 对称二叉树</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#104-%E4%BA%8C%E5%8F%89%E6%A0%91%E7%9A%84%E6%9C%80%E5%A4%A7%E6%B7%B1%E5%BA%A6"><span class="toc-number">3.0.4.5.</span> <span class="toc-text">104. 二叉树的最大深度</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#110-%E5%B9%B3%E8%A1%A1%E4%BA%8C%E5%8F%89%E6%A0%91"><span class="toc-number">3.0.4.6.</span> <span class="toc-text">110. 平衡二叉树</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#226-%E7%BF%BB%E8%BD%AC%E4%BA%8C%E5%8F%89%E6%A0%91"><span class="toc-number">3.0.4.7.</span> <span class="toc-text">226. 翻转二叉树</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#98-%E9%AA%8C%E8%AF%81%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91"><span class="toc-number">3.0.4.8.</span> <span class="toc-text">98. 验证二叉搜索树</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E4%BA%8C%E5%88%86%E6%9F%A5%E6%89%BE%EF%BC%8810%EF%BC%89"><span class="toc-number">3.0.5.</span> <span class="toc-text">二分查找（10）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#69-X-%E7%9A%84%E5%B9%B3%E6%96%B9%E6%A0%B9"><span class="toc-number">3.0.5.1.</span> <span class="toc-text">69. X 的平方根</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#35-%E6%90%9C%E7%B4%A2%E6%8F%92%E5%85%A5%E4%BD%8D%E7%BD%AE"><span class="toc-number">3.0.5.2.</span> <span class="toc-text">35. 搜索插入位置</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#35-%E5%9C%A8%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E6%9F%A5%E6%89%BE%E5%85%83%E7%B4%A0%E7%9A%84%E7%AC%AC%E4%B8%80%E4%B8%AA%E5%92%8C%E6%9C%80%E5%90%8E%E4%B8%80%E4%B8%AA%E4%BD%8D%E7%BD%AE"><span class="toc-number">3.0.5.3.</span> <span class="toc-text">35. 在排序数组中查找元素的第一个和最后一个位置</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#74-%E6%90%9C%E7%B4%A2%E4%BA%8C%E7%BB%B4%E7%9F%A9%E9%98%B5"><span class="toc-number">3.0.5.4.</span> <span class="toc-text">74. 搜索二维矩阵</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#153-%E5%AF%BB%E6%89%BE%E6%97%8B%E8%BD%AC%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC"><span class="toc-number">3.0.5.5.</span> <span class="toc-text">153. 寻找旋转排序数组中的最小值</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#33-%E6%90%9C%E7%B4%A2%E6%97%8B%E8%BD%AC%E6%8E%92%E5%BA%8F%E6%95%B0%E7%BB%84"><span class="toc-number">3.0.5.6.</span> <span class="toc-text">33. 搜索旋转排序数组</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#278-%E7%AC%AC%E4%B8%80%E4%B8%AA%E9%94%99%E8%AF%AF%E7%89%88%E6%9C%AC"><span class="toc-number">3.0.5.7.</span> <span class="toc-text">278. 第一个错误版本</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#162-%E5%AF%BB%E6%89%BE%E5%B3%B0%E5%80%BC"><span class="toc-number">3.0.5.8.</span> <span class="toc-text">162. 寻找峰值</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#287-%E5%AF%BB%E6%89%BE%E9%87%8D%E5%A4%8D%E6%95%B0"><span class="toc-number">3.0.5.9.</span> <span class="toc-text">287.寻找重复数</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#275-H-%E6%8C%87%E6%95%B0%E2%85%A1"><span class="toc-number">3.0.5.10.</span> <span class="toc-text">275. H 指数Ⅱ</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E5%AD%97%E7%AC%A6%E4%B8%B2%EF%BC%889%EF%BC%89"><span class="toc-number">3.0.6.</span> <span class="toc-text">字符串（9）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#38-%E5%A4%96%E8%A7%82%E6%95%B0%E5%88%97"><span class="toc-number">3.0.6.1.</span> <span class="toc-text">38. 外观数列</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#49-%E5%AD%97%E6%AF%8D%E5%BC%82%E4%BD%8D%E8%AF%8D%E5%88%86%E7%BB%84"><span class="toc-number">3.0.6.2.</span> <span class="toc-text">49. 字母异位词分组</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#191-%E5%8F%8D%E8%BD%AC%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B8%AD%E7%9A%84%E5%8D%95%E8%AF%8D"><span class="toc-number">3.0.6.3.</span> <span class="toc-text">191. 反转字符串中的单词</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#165-%E6%AF%94%E8%BE%83%E7%89%88%E6%9C%AC%E5%8F%B7"><span class="toc-number">3.0.6.4.</span> <span class="toc-text">165. 比较版本号</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#929-%E7%8B%AC%E7%89%B9%E7%9A%84%E7%94%B5%E5%AD%90%E9%82%AE%E4%BB%B6%E5%9C%B0%E5%9D%80"><span class="toc-number">3.0.6.5.</span> <span class="toc-text">929. 独特的电子邮件地址</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#3-%E6%97%A0%E9%87%8D%E5%A4%8D%E5%AD%97%E7%AC%A6%E7%9A%84%E6%9C%80%E9%95%BF%E5%AD%90%E4%B8%B2"><span class="toc-number">3.0.6.6.</span> <span class="toc-text">3. 无重复字符的最长子串</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#5-%E6%9C%80%E9%95%BF%E5%9B%9E%E6%96%87%E4%B8%B2"><span class="toc-number">3.0.6.7.</span> <span class="toc-text">5. 最长回文串</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#208-%E5%AE%9E%E7%8E%B0-trie-%E5%89%8D%E7%BC%80%E6%A0%91"><span class="toc-number">3.0.6.8.</span> <span class="toc-text">208. 实现 trie 前缀树</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#8-%E5%AD%97%E7%AC%A6%E4%B8%B2%E8%BD%AC%E6%8D%A2%E6%95%B4%E6%95%B0-atoi"><span class="toc-number">3.0.6.9.</span> <span class="toc-text">8. 字符串转换整数 - atoi</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E5%8F%8C%E6%8C%87%E9%92%88%EF%BC%882%EF%BC%89"><span class="toc-number">3.0.7.</span> <span class="toc-text">双指针（2）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#167-%E4%B8%A4%E6%95%B0%E4%B9%8B%E5%92%8C-ii-%E8%BE%93%E5%85%A5%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84"><span class="toc-number">3.0.7.1.</span> <span class="toc-text">167. 两数之和-ii-输入有序数组</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#26-%E5%88%A0%E9%99%A4%E6%9C%89%E5%BA%8F%E6%95%B0%E7%BB%84%E4%B8%AD%E7%9A%84%E9%87%8D%E5%A4%8D%E9%A1%B9"><span class="toc-number">3.0.7.2.</span> <span class="toc-text">26. 删除有序数组中的重复项</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#%E6%BB%91%E5%8A%A8%E7%AA%97%E5%8F%A3%EF%BC%881%EF%BC%89"><span class="toc-number">3.0.8.</span> <span class="toc-text">滑动窗口（1）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#76-%E6%9C%80%E5%B0%8F%E8%A6%86%E7%9B%96%E5%AD%90%E4%B8%B2"><span class="toc-number">3.0.8.1.</span> <span class="toc-text">76. 最小覆盖子串</span></a></li></ol></li><li class="toc-item toc-level-3"><a class="toc-link" href="#DFS%EF%BC%88%E6%B7%B1%E5%BA%A6%E6%90%9C%E7%B4%A2%EF%BC%89-%E5%9B%9E%E6%BA%AF%EF%BC%884%EF%BC%89"><span class="toc-number">3.0.9.</span> <span class="toc-text">DFS（深度搜索） + 回溯（4）</span></a><ol class="toc-child"><li class="toc-item toc-level-4"><a class="toc-link" href="#17-%E7%94%B5%E8%AF%9D%E5%8F%B7%E7%A0%81%E7%9A%84%E5%AD%97%E6%AF%8D%E7%BB%84%E5%90%88"><span class="toc-number">3.0.9.1.</span> <span class="toc-text">17. 电话号码的字母组合</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#79-%E5%8D%95%E8%AF%8D%E6%90%9C%E7%B4%A2"><span class="toc-number">3.0.9.2.</span> <span class="toc-text">79. 单词搜索</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#46-%E5%85%A8%E6%8E%92%E5%88%97"><span class="toc-number">3.0.9.3.</span> <span class="toc-text">46. 全排列</span></a></li><li class="toc-item toc-level-4"><a class="toc-link" href="#47-%E5%85%A8%E6%8E%92%E5%88%97-%E2%85%A1"><span class="toc-number">3.0.9.4.</span> <span class="toc-text">47. 全排列 Ⅱ</span></a></li></ol></li></ol></li></ol></li></ol>
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